Quadratic Factorization
A reasonable mastery of polynomial factorization will facilitate the understanding of certain kinds of mathematics, certain later proofs, and higher scores on popular standardized tests. We will first discuss the simpler cases of quadratic factoring and build from these principles to discuss higher degree polynomial factorization techniques.
0. Tips for Studying and Summary Material for Notecards
You’ll hear and see this piece of advice often: make notecards of information that you are trying to rapidly recall. The action of writing them down helps to inculcate them in the brain. Put essential notecards on your bathroom mirror and review them before bed and when you wake up. You will learn them more quickly and with less effort this way.
- $(a+b)(a-b) = a^2 – b^2$
- $(a \pm b)^2 = a^2 \pm 2ab +b^2$
A big portion of the purpose of this section is to help you appreciate what you will learn in calculus. Factoring polynomials with the tools of only algebra is extremely difficult and does not always even work. We cannot guaranty that a polynomial has rational or real roots. Each tool you find in this section is special and unique. One is like a Philip’s head screwdriver, the next a flathead screwdriver, and the next a hex wrench. This will help you to appreciate Newton’s Method from calculus when you get to it. It is like a versatile power drill. You will be able to factor arbitrary polynomials to whatever precision you care for.
1. Factoring Polynomials: The Basics
Factoring a polynomial is in some sense like reverse engineering (or working backwards) through the polynomial distribution process. When we factor a polynomial, we are asking “what were the polynomials (if any) that were multiplied together to get this polynomial I am currently looking at?” Why might we ask such a question? Well there are some important aspects of a polynomial that are best understood by knowing the “factors” of that polynomial. A factor of a polynomial $P(x)$ is any polynomial (often or ideally a binomial) that evenly divides into $P(x)$. Just like the number 3 is a factor of the number 12, so too is the polynomial (binomial) $(x-2)$ a factor of the polynomial $x^2-4x+4$.
Since polynomials come in infinite varieties and degrees and since they can potentially factor into polynomials of infinite varieties and degrees, there are many methods to factor polynomials and very often a polynomial cannot be factored completely or at all. Think about the analogy of numerical factoring: some numbers like the prime numbers cannot be factored into simpler quantities; for example, 113 is as simple as this number potentially gets, since it cannot be factored into smaller prime factors; 113 is a prime number itself. However, 112 can be factored into (66)(2) and then (33)(2)(2) and then (11)(3)(2)(2). Polynomials work in a similar way; however, we sometimes have to do some clever work to find the factors of a polynomial. In this chapter you will encounter all the standard methods to factor polynomials.
Generally speaking, it is more difficult to factor polynomials the higher their degree. The higher the degree of a polynomial the more we might need to rely on computational methods (i.e. algorithms best executed by a computer) rather than analytical solutions. However, given quadratic polynomials (i.e second degree polynomials) and cubic polynomials (i.e. third degree polynomials) there are many ways to potentially find solutions. We will examine these nice cases first.
Quadratic Polynomials: The Anatomy of Parabolas and Their Functions
Just like lines had various forms to their function notation (i.e. slope-intercept, point-slope, and quadratic functions are written in three forms: standard form, factored form and vertex form. We will talk about vertex form in a later section; for now and in most cases, we will consider quadratics in standard form and factored form:
Quadratic Standard Form: $y=ax^2+bx+c$ where $a$, $b$, and $c$ are constant coefficients and $a\ne{0}$. For example $y=x^2+2x$ has coefficient values $a=1$, $b=2$, and $c=0$ while $y=x^2-5$ has coefficient values $a=1$, $b=0$, and $c=-5$
Quadratic Standard (“Factored”) Form: $y=(Ax\pm{B})(Cx\pm{D})$ where $A$, $B$, $C$, and $D$ are constant coefficients with $A\ne{0}$ and $C\ne{0}$. For example, $y=(x+4)(x-2)$ has coefficient values $A=1$, $B=4$, $C=1$, and $D=-2$, while $y=x(x\sqrt{2}-\pi)$ has coefficient values $A=1$, $B=0$, $C=\sqrt{2}$, and $D=-\pi$.
The goal of the prior unit on polynomial distribution was to move us from the factored form of a polynomial to its standard form. The goal of this unit is to undo this distribution process and find the factored form that generated the standard form (should such a factored form exist at all).
Before we delve deeper into the topic of factoring, we first need to have a richer discussion on the nature of polynomials in terms of the x-y plane. We have discussed polynomials in terms of monomial simplification, but we have not yet talked about their graphs as functions. Understanding these graphs will help us to better understand the meaning and benefits of factoring a polynomial.
Consider the polynomials to the right; these graphs are of the quadratic polynomials $x^2+2x-8$, $-x^2-x+6$, and $\frac{1}{3}x^2+x+2$. What are the important anatomical parts to these graphs? Well, when we were considering linear functions, we found that the slope and y-intercept were the most important parts of any line, determining its entire nature. We have some similarly important parts of this “parabola,” which is the name we give to the shape of a quadratic polynomial; you can read more about this naming convention under the Conic Sections portion of the chapter on Implicit Functions.
Parabolas can also have y-intercepts and x-intercepts; however, as in the case of the graph of $\frac{1}{3}x^2+x+2$, we might have no x-intercepts. Thus, there is no guarantee, given any randomly generated parabola, that it will have x-intercepts. The x-intercepts of any polynomial, quadratic or otherwise, are often called the roots of the polynomial or solutions to the polynomial. These concepts will become important shortly.
Though we have no guarantee of x-intercepts, we must always have a y-intercept, since the graph of a parabola extends to infinity in the left and right direction (i.e. the “domain of the quadratic function is defined for all Real Number values,” meaning that there is no value for $x$ such that the function will evaluate as “undefined”).
Furthermore, given the interesting shape of the parabolas, we can also see a minimum or maximum y-value at the tip of its curve, which we call the vertex of the parabola. This value is a minimum when the parabola opens upwards and a maximum when the parabola opens downwards. A parabola’s direction is determined by the highest degree term’s learning coefficient: if this term is positive ($ax^2$) the parabola will open up; if this term is negative $-ax^2$ the parabola will open down. The vertex determines the range of the quadratic function, since it establishes a lower or upper bound of possible y-values. Thus, to know the range of our quadratic function, we must know its vertex and direction.
It is also worth noting that quadratic functions are symmetric about some vertical axis, which is often called the axis of symmetry; if we draw a vertical line through the minimum or maximum point, the two sides of the parabola will be graphically symmetric.
This is all summed up in the image below. Learn this anatomy and vocabulary, since it will be crucial in our discussion of quadratics and other polynomials.
Now that we have looked at a selection of quadratics, let us take a step back and consider the most simple quadratic function: $y=x^2$. This is called the parent function of the quadratic functions. Every function has a parent function, and we will discuss the parent function for each type of function as it arises. It is useful to think of any class of function according to transformations of its parent function; furthermore, most of the fundamentals aspects of a class of functions can be observed by observing the parent function.
This function has one x-intercept and one y-intercept, which is the same point at the origin. It is symmetric about the y-axis or $x=0$; we can observe the nature of this symmetry with the plotted points: given some x-value and its corresponding negative, the y-values will be the same (i.e. given $f(x)=x^2$, $f(x)=f(-x)$ for all $x$ in $\mathbb{R}$). Essentially, this tells us that if we know where the axis of symmetry is, we can determine all y-values for both sides given only one side of the parabola.
How do we know where $y=x^2$ crosses the y-axis? Well, just like in lines, the y-intercept happens where $x=0$. Thus, we can quickly find the y-intercept for any parabola by simply plugging in the value $x=0$ into the function and evaluating. For this function $f(x)=x^2$, $f(0)=0^2=0 and so (0, 0) is our y-intercept.
The x-intercept for this graph is also clearly (0, 0), but how do we frame this generally? Well, the x-intercept of any function happens when $y=0$. Thus, to know where the function crosses the x-axis, we must ask the question of when is $y=0$ for the function? This might be an easy question, such as with $y=x^2$ (i.e. “when does 0=x^2?” Well, when $x=0$ of course!); however, for more complex parabolic functions this question might not have such an obvious answer (i.e. given $y=3x^2-4x-7$, “when does $0=3x^2-4x-7$ ?”).
Given any random polynomial, quadratic or otherwise, when does the function evaluate to zero (i.e. “When is $P(x)=0$ for some polynomial $P(x)$”)? It is this kind of difficult question that the methods of factoring help us to answer.
2. Quadratic Polynomials: Factoring with Rational Solutions
If we are lucky, our quadratic polynomial will have rational solutions; if we are really lucky, our quadratic polynomial will have integer solutions. This section will show you how to tackle these nice instances of rational solutions to quadratics.
Let us begin by considering a general case of an already factored polynomial:
$$(x+A)(x+B)$$
If we distribute this polynomial, we will return:
$$x^2+Bx+Ax+AB$$
Now let us “combine” the like terms on $x$ by factoring out the $x$ from both of them.
$$x^2+(A+B)x+AB$$
Assuming $A$ and $B$ are integers, we can take this form as a map to reversing the distribution operation we have done. Consider the following:
If there is a quadratic polynomial, $P(x)$, in the form $P(x)=x^2+Nx+M$ and $N$ and $M$ are integers, then the polynomial is potentially factorizable into the form $P(x)=(x+A)(x+B)$ if $A$ and $B$ are factors of $M$ and sum to equal $N$. That is $M=A(B)$ and $N=A+B$ if $P(x)=x^2+Nx+M=(x+A)(x+B).
Consider the following example: (click to expand)
Consider the following example: $x^2+2x-15$ (click to expand)
$x^2+2x-15$
The factors of -15 are $\pm5$, $\pm3$, $\pm1$, and $\pm15$
The only two of these factors that will both multiply to equal -15 and add to equal +2 are
$$M=(5)(-3)=-15\quad\text{and}\quad{N=5+-3=2}$$
Thus, $x^2+2x-15=(x+5)(x-3)$
Since multiplication is commutative, it does not matter in what order you list these factors. That is,
$$(x+5)(x-3)=(x-3)(x+5)$4
To find the “solutions” or “zeros” to this polynomial, we need to ask “where does this cross the x-axis,” which is the same as asking “what are the values of $x$ when we set $y=0$ ?” So, we should set this factored polynomial equal to zero by substituting zero for $y$.
$$y=(x+5)(x-3)\rightarrow{0=(x+5)(x-3)}$$
By the Multiplicative Property of Zero, we know that if $0=ab$ for some values $a$ and $b$ (which can be numbers or whole expressions), then either $a$, $b$, or both values must be zero. Thus, if $0=(x+5)(x-3)$, then we can assume that either $0=(x+5)$ or $0=(x-3)$. Now we should ask what the value of $x$ must be for either factor to be zero:
$$0=x+5\quad\quad0=x-3$$
$$0-5=x+5-5\quad\quad0+3=x-3+3$$
$$-5=x\quad\quad3=x$$
Thus, the parabola crosses $y=0$ (namely, the x-axis) when either $x=3$ or when $x=05$. Let us test this answer by substituting the values of $x$ back into the factored and unfactored polynomial:
$$P(x)=x^2+2x-15=(x+5)(x-3)$$
$$P(-5)={-5}^2+2(-5)-15=(-5+5)(-5-3)\quad\quad{P(3)=3^2+2(3)-15=(3+5)(3-3)}$$
$$P(-5)=25+{-10}-15=(0)(-8)\quad\quad{P(3)=9+6-15=(8)(0)}$$
$$P(-5)=25+{-25}=(0)(-8)\quad\quad{P(3)=15-15=(8)(0)}$$
$$P(-5)=0=0\quad\quad{P(3)=0=0}$$
The graph of the function confirms our work as well:
Try these next problems on your own. Click the problem to see the solution.
Factor the polynomial $y=x^2-10x+24$ and give its solutions (show solution)
The factors of 24 are $\pm1$ and $\pm24$; $\pm4$ and $\pm6$; $\pm3$ and $\pm8$; and $\pm2$ and $\pm12$. The only set of two factors that can be summed in some fashion to equal -10 are $\pm4$ and $\pm6$.
$$M=24=(-6)(-4)\quad\quad{N=-6+-4=-10}$$
Thus, $y=x^2-10x+24=(x-6)(x-4)$
If we consider where $y=0$, then
$$0=(x-6)(x-4)\rightarrow\text{implied that}\rightarrow{x-6=0\quad\text{or}\quad{x-4=0}}$$
Thus,
x-6=0\rightarrow{x=6}
x-4=0\rightarrow{x=4}
We can confirm this with the graph of the function:
Factor the polynomial $y=x^2+6x+9$ and give its solutions (show solution)
The factors of 9 are $\pm1$ and $\pm9$ as well as $\pm3$ and $\pm3$. The only set of two factors that can be summed in some fashion to equal 6 are $\pm3$ and $\pm3$.
$$M=9=(3)(3)\quad\quad{N=3+3=-6}$$
Thus, $y=x^2-10x+24=(x+3)(x+3)$
If we consider where $y=0$, then
$$0=(x+3)(x+3)\rightarrow\text{implied that}\rightarrow{x+3=0\quad\text{or}\quad{x+3=0}}$$
Thus,
x+3=0\rightarrow{x=-3}
x+3=0\rightarrow{x=-3}
But wait! These are the same solutions. This merely means that the parabola only hits the x-axis one time; the only way this is possible for a parabola is if the tip of the parabola touches the x-axis.We can confirm this with the graph of the function:
This is a perfect square polynomial. These are somewhat common, and the general form of the perfect square polynomial should be remembered. If you see a polynomial in the form $y=x^2\pm{2bx}+b^2$, then it will always factor into the form $(x\pm{b})(x\pm{b})=(x\pm{b})^2$.
$$y=x^2\pm{2bx}+b^2=(x\pm{b})(x\pm{b})=(x\pm{b})^2$$
Factor the polynomial $y=x^2-16$ and give its solutions (show solution)
Note that the value of $N=0$
The factors of 16 are $\pm1$ and $\pm16$; $\pm2$ and $\pm8$; as well as $\pm4$ and $\pm4$. The only set of two factors that can be summed in some fashion to equal 0 are $4$ and $-4$.
$$M=-16=(-4)(4)\quad\quad{N=-4+4=0}$$
Thus, $y=x^2-16=(x-4)(x+4)$
If we consider where $y=0$, then
$$0=(x-4)(x+4)\rightarrow\text{implies that}\rightarrow{x-4=0\quad\text{or}\quad{x+4=0}}$$
Thus,
$$x-4=0\rightarrow{x=4}$$
$$x+4=0\rightarrow{x=-4}$$
We can confirm this with the graph of the function:
This is a particularly nice parabola, since it is symmetric about the y-axis. This is merely because we have taken the parent function of the parabola (i.e. $y=x^2$) and shifted it down 16 units (ergo the minus 16 in $y=x^2-16$).
This kind of parabola is called the “difference of perfect squares” polynomial. These are somewhat common, and the general form of the perfect square polynomial should be remembered. If you see a polynomial in the form $y=x^2-b^2$, then it will always factor into the form $(x-b)(x+b)$.
$$y=x^2-b^2=(x-b)(x+b)$$
We can verify that this is a variation of the parent quadratic function $y=x^2$ by considering that $b^2=0$ is implied in the parent function:
$$y=x^2=x^2-0^2=(x+0)(x-0)=(x)(x)=x^2$$
Let us consider an even more general case, where all monomials have coefficients:
$$(Cx+A)(Dx+B)$$
If we distribute this polynomial, we will return:
$$CDx^2+BCx+ADx+AB$$
Now let us “combine” the like terms on $x$ by factoring out the $x$ from both of them.
$$CDx^2+(AD+BC)x+AB$$
Assuming $A$, $B$, $C$, and $D$ are integers, we can take this form as a map to reversing the distribution operation we have done. Consider the following:
If there is a quadratic polynomial, $P(x)$, in the form $P(x)=Kx^2+Nx+M$ and $N$, $M$, and $K$ are integers, then the polynomial is potentially factorizable into the form $P(x)=(Cx+A)(Dx+B)$ if
- $A$ and $B$ are factors of $M$
- $C$ and $D$ are factors of $K$
- And the sum $AD+BC=N$. That is $M=A(B)$ and $N=A+B$ if $P(x)=Kx^2+Nx+M=CDx^2+(AD+BC)x+AB=(Cx+A)(Dx+B).
Consider the following example: $P(x)=3x^2+2x-8$ (click to expand)
The factors of $K$ are only 3 and 1, which makes our job a bit easier. We will need to consider these factors with $M$. The factors of $M$ are $\pm1$ and $\pm8$ as well as $\pm2$ and $\pm4$.
Since $N=2$, let us consider the sum of factor products that equal 2:
$$AD+BC=2$$
This might take some trial and error, but we know that $C$ and $D$ must be either 3 and 1 or 1 and 3.
We might try ever combination, but you might also see the more likely solution path through intuition:
$$A(D)+B(C)=N$$
$$8(3)\pm1(1)=25\text{and}\,23$$
$$1(3)\pm8(1)=11\text{and}\,-5$$
$$2(3)\pm4(1)=10\text{and}\,2$$
$$4(3)\pm2(1)=14\text{and}\,10$$
It appears that $A(D)+B(C)=2(3)-4(1)=2$ is our right path.
Thus
$$A=2$$
$$B=-4$$
$$C=1$$
$$D=3$$
And so
$$(Cx+A)(Dx+B)=(x+2)(3x-4)$$
Thus, setting our polynomial equal to zero, we can find the x-intercepts via the Multiplication Property of Zero.
$$y=(x+2)(3x-4)\rightarrow0=(x+2)(3x-4)$$
$$0=(x+2)(3x-4)\rightarrow\text{implies that}\rightarrow{x+2=0\quad\text{or}\quad{3x-4=0}}$$
Thus,
$$3x-4=0\rightarrow{3x=4}\rightarrow{x=\frac{4}{3}}$$
$$x+2=0\rightarrow{x=-2}$$
We can confirm this with the graph of the function:
Try these next problems on your own. Click the problem to see the solution.
Factor the polynomial $5x^2-17x+6$ and give its solutions (show solution)
The factors of $K$ are only 5 and 1, which makes our job a bit easier. We will need to consider these factors with $M$. The factors of $M$ are 1 and 6 as well as 2 and 3.
Since $N=-17$, let us consider the sum of factor products that equal -17:
$$AD+BC=-17$$
This might take some trial and error, but we know that $C$ and $D$ must be either 5 and 1 or 1 and 5.
We might try ever combination, but you might also see the more likely solution path through intuition:
$$A(D)+B(C)=N$$
$$2(5)+3(1)=13$$
$$3(5)+2(1)=17$$
$$1(5)+6(1)=11$$
$$6(5)+1(1)=31$$
Since AD+BC=+17 for $A=5$, $B=1$, $C=3$, $D=2$, we know that these are the general values that we need. However, we are looking for $N=-17$ not $N=+17$; thus, we need to make $C$ and $D$ both negative (i.e. $K=5=CD=-5(-1)$) or $A$ and $B$ both negative (i.e. $M=6=AB=-3(-3)$).
Ultimately, our choice will not matter in terms of our solutions; however, for ease of reading and writing, the constants are better to make negative than the coefficients of $x$; thus let us make $A$ and $B$ negative. Consequently, $A=-3$, $B=-2$, $C=1$, and $D=5$, and so our factored form is:
$$(Cx+A)(Dx+B)=(x-3)(5x-2)$$
Thus, setting our polynomial equal to zero, we can find the x-intercepts via the Multiplication Property of Zero.
$$y=(x-3)(5x-2)\rightarrow0=(x-3)(5x-2)$$
$$0=(x-3)(5x-2)\rightarrow\text{implies that}\rightarrow{5x-2=0\quad\text{or}\quad{x-3=0}}$$
Thus,
$$5x-2=0\rightarrow{5x=2}\rightarrow{x=\frac{2}{5}}$$
$$x-3=0\rightarrow{x=3}$$
We can confirm this with the graph of the function:
Factor the polynomial $4x^2+10x-6$ and give its solutions (show solution)
First, it will be best to factor out the common constant factor of $K$, $M$, and $N$, which is 2.
$$4x^2+10x-6\rightarrow{2(2x^2+5x-3)}$$
Now we are considering the polynomial $2x^2+5x-3$
$K$ has only the factors 2 and 1. $M$ has the factors $\pm1$ and $\mp3$. Now let us consider $N=5$ knowing that $N=AD+BC$
$$A(D)+B(C)=N$$
$$1(2)+(-3)(1)=-1$$
$$(-3)(2)+(1)(1)=-5$$
$$(-1)(2)+(3)(1)=1$$
$$(3)(2)+(-1)(1)=5$$
Thus, $A(D)+B(C)=(3)(2)+(-1)(1)=5=N$
$$A=3$$
$$B=-1$$
$$C=1$$
$$D=2$$
And so $y=2x^2+5x-3=(Cx+A)(Dx+B)=(x+3)(2x-1)$$
Thus, setting our polynomial equal to zero, we can find the x-intercepts via the Multiplication Property of Zero.
$$y=(x+3)(2x-1)\rightarrow0=(x+3)(2x-1)$$
$$0=(x+3)(2x-1)\rightarrow\text{implies that}\rightarrow{x+3=0\quad\text{or}\quad{2x-1=0}}$$
Thus,
$$2x-1=0\rightarrow{2x=1}\rightarrow{x=\frac{1}{2}}$$
$$x+3=0\rightarrow{x=-3}$$
These solutions hold for both the polynomial with the 2 factored out as well as the original polynomial.
We can confirm this with the graph of both functions:
$$f(x)=4x^2+10x-6=2(x+3)(2x-1)$$
$$g(x)=2x^2+5x-3=(x+3)(2x-1)$$
Note, however, that both functions are quite different, and so we must be sure to include the factored out 2 in our final statement of the factored polynomial, $P(x)=2(x+3)(2x-1)$
Obviously, this could be much, much more difficult to determine intuitively than the simpler case of $(x+A)(x+B)$; thus, we generally might want a better method to tackle quadratics with coefficients on the leading monomial. Furthermore, all of these quadratic polynomials are pedagogical and not likely to model real-world problems. More often than not, a quadratic will not have rational solutions; they might have irrational or even imaginary solutions. To tackle these instances, we need a few more tools in our factoring toolbelt.
3. Essential Rapid Factoring and Distribution Techniques
As seen in some of the prior examples, there are a few special factoring forms. These are worth knowing from memory and readily recognizing. Be sure to dedicate these to your memory.
Another important feature to pay careful attention to is the fact that equalities are symmetrical. It is advantageous to be able to go from the factored for to the distributed form quickly and fluidly.
$\begin{align*}
\tag{1} a^2 \pm 2ab + b^2 & = (a \pm b)^2 \\
\tag{2} a^2 – b^2 & = (a + b)(a – b) \\
\tag{3} a^n – b^n & = (a – b)(a^{n-1} + ba^{n-2} + \cdots + b^{n-2}a + b^{n-1})
\end{align*}$
Equation 1 from above is the sum or difference of squares. Equation 2 is the difference of perfect squares. Equation 3 is a little less used in lower mathematics, but frequently used in higher mathematics. Equation 3 is related to both the difference of perfect squares (n = 2) and general geometric series so we will call it a geometric difference. Equations 1 and 2, we can verify by hand, but equation 3 will require a proof.
Note that we consider $n \ge 2$ and then
$$\begin{align*}
a^n – b^n &= a^n – b^n + (a^{n-1}b + \cdots + ab^{n-1}) – (a^{n-1}b + \cdots + ab^{n-1}) \\
&= a^n + (a^{n-1}b + \cdots + ab^{n-1}) – b^n – (a^{n-1}b + \cdots + ab^{n-1}) \\
&= a(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) – b(b^{n-1} + a^{n-1} + \cdots + ab^{n-2}) \\
&= (a – b)(a^{n-1} + (a^{n-2}b + \cdots + b^{n-1}))
\end{align*}$$
A fourth essential technique involves recognize the greatest common factor. The greatest common factor is a term that appears across a collection of terms and can be factored out. For instance $4x^2 + 6x + 12x^3$ has the greatest common factor $2x$ and can be rewritten as $2x(2x + 3 + 6x^2$. Incidentally, it is common to order polynomial expressions by ascending or descending degrees and we’ll try to keep to this practice. In that case, the above becomes $2x(6x^2 + 2x + 3)$ and this is the most common way a solution would be written.
4. Quadratic Polynomials: Completing the Square and Vertex Form
Recall what we learned from Function Transformations; we can use some of the concepts from that unit to understand, in a more compact and generalized form, the many varieties of parabolic functions we might encounter. Let us begin by examining the parent function of quadratics: $y=x^2$.
What happens to the graph of the function when we add a constant inside the exponentiation operations, such as in $y=(x+a)^2$. Does this look at all familiar? It should!
$y=(x+a)^2$ is a perfect square polynomial with repeated factors $(x+a)$. When a factor repeats, the parabola touches, but does not cross the x-axis. Consider the graph of $y=(x+3)^2$. Visually, we have moved the parent function 3 units to the left by adding 3 units “inside the parent function” of the square exponent (i.e $f(x+3)=x^2$). This seems somewhat counterintuitive: adding 3 to move to the left 3. Why does this happen?
Well, we have relocated the minimum value (vertex of the parabola). While the parent function evaluates with a minimum value at $f(0)=0$, $g(x)$, to have its minimum must shift to the left 3 units to compensate for the addition of 3 to the right within the parent function; that is, $g(x)=f(x+3)=(x+3)^3$ has its minimum at -3 because $f(x+3)$ evaluated at -3 is $f(-3+3)=f(0)=0$.
This might be a bit confusing. Understanding will come with thought and practice; however, a general rule may be remembered and applied:
If adding within the parent function, move to the left; if subtracting (adding a negative) in the parent function, move to the right.
Next, what happens when we multiply the parent function by some constant value: $y=a(x^2)$. This just takes all y-values from the parent function and increases them by a constant rate: where $y=4$, we will now have $a(4); where $y=25$, we will now have $a(25); and so forth.
Thus, if $a>1$ then we are stretching the parabola along the y-axis; if $0<a<1$, then we are compressing the parabola along the y-axis; if $a<0$ then the rules for positive $a$-values still apply, but the parabola will be reflected across the x-axis. Consider the graphical examples below:
Last, what happens to the parent function if we merely add a value to it outside the parent function: $y=x^2+a$ ? The result is simple enough! Just like with a linear function,
adding to the parent function itself merely changes the y-intercept, mapping all y-values of the parent function to new values at a constant distance up or down the y-axis.
Interesting to note is that in this form, we are not only moving the y-intercept; the y-intercept itself is the vertex of the parabola. Thus, we can move the vertex up or down by adding or subtracting from the parent function.
Thus, we can summarize these observations as follows:
- $y=(x+a)^2$ moves the vertex to the left or the right depending on $a$ ($a>0$ moves left and $a<0$ moves right)
- $y=x^2+a$ moves the vertex of the parabola up or down depending on $a$ ($a>0$ moves up and $a<0$ moves down).
- $y=a(x^2)$ stretches or compresses the parabola ($a>1$ stretches, $0<a<1$ compresses, $a<0$ flips all relationships and compresses or stretches depending on the negative magnitude of $a$).
If we synthesize all these potential transformations into one general function we can say that a parabola can be modeled by the relocation of its vertex and dilation of its values:
$$y=a(x-h)^2+k$$
Where $a$ dilates the function according to the third rule above and $h$ and $k$ determine the vertical distance $y=k$ and horizontal distance $x=h$ the vertex has been relocated from the vertex of the parent function at the origin.
$y=a(x-h)^2+k$ is consequently called the vertex form of the parabolic parent function (i.e. quadratic function); it has embedded within it the coordinate of the vertex: $(h, k)$. Note that the value of $h$ in the coordinate does not account for the subtraction in the parent function (i.e. if $h$ is being subtracted, the value of $h$ in the vertex is $h$, not $-h$. Consequently, if we have a vertex form like $y=a(x+h)^2+k$, then the value of $h$ in the vertex coordinate must be $-h$, since $a(x+h)^2+k=a(x-{-h})^2+k$.
There are some useful things we can know about the parabola if we know the vertex point.
- First, we know that the parabola is symmetric on either side of the vertex.
- Second, depending on whether $a$ is positive or negative, we know the maximum or minimum of the parabolic function.
- Third, if we know one factor of the function, we know the other factor is symmetric about the vertex and can easily locate it consequently.
There are some weakness to vertex form as compared to the factored and standard form:
- The vertex form does not tell us anything about the location of the parabola’s factors/zeros.
- We cannot factor the vertex form without first expanding it out and putting it back into standard form.
We can certainly put the vertex form back into standard form and then factor it. Consider the below example:
Example: Factor $y=2(x-3)^2-8$ (click to expand)
To factor, we should first put the polynomial in standard form through expanding and simplifying it. Note that the vertex form always has a perfect square polynomial in it. This will be important to keep in mind now and in a bit. Certainly, recognizing the perfect square polynomial helps to rapidly distribute the polynomial.
$$2(x-3)^2-8$$
$$2(x-3)(x-3)-8$$
$$2(x^2-6x+9)-8$$
$$2x^2-12x+18-8$$
$$2x^2-12x+10$$
Now let us factor this using the methods we have learned thus far:
First, we should factor any common constant factor, which in the case of this polynomial is 2:
$$2x^2-12x+10\rightarrow2(x^2-6x+5)$$
Now let us factor according to $x^2+Nx+M=(X+A)(X+B)$ where $M=AB$ has factors $A=\pm1$ and $B=\pm5$. These factors also need to add to equal -6 (i.e. N=A+B=-6$).
$-1+-5=-6\quad\text{and}\quad(-1)(-5)=5$
Thus
$$(x+A)(x+B)\rightarrow(x-5)(x-1)$$
And so, the parabola has x-intercepts at $x=5$ and $x=1$ and a factored form of $p(x)=2(x-5)(x-1)$. Don’t forget the 2 we factored out at the outset!
We can use these factors to confirm our original vertex by knowing that the x-value of the vertex should lie perfectly between the two x-intercepts which are symmetric about the x-value of the vertex. Thus, what is the x-value directly between $x=5$ and $x=1$? This should be $x=3$. The function evaluated at $x=3$ is
$$p(x)=2(x-5)(x-1)\rightarrow{p(3)=2(3-5)(3-1)=2(-2)(2)=-8}$
Given our original function in vertex form – $y=2(x-3)^2-8$ – we can see that the vertex $(h, k)$ is $(3, -8)$. Thus, our work is proven to be consistent.
So, what if we want the vertex form when we have the standard form of a quadratic? Well, what we need is to derive the perfect square polynomial that is in the vertex form of the parabola. To do this, we will “complete the square;” that is, we will add what is necessary to the function to make it a perfect square polynomial with whatever extra is needed to balance the addition appended to the end. Consider this geometric example:
The area of a square with side lengths $x$ is added to the area of a rectangle with side lengths $b$ and $x$. If these areas were repartitioned to form a square, what would be the area of the missing part necessary to complete the square?
First, if we evenly divide the rectangle along its width $b$ we would get two rectangles with side lengths $x$ and $\frac{b}{2}$; we can then place these two rectangles along adjacent sides of the square so that the lengths $x$ of each rectangle coincide with the $x$ sides of the square, as seen in the middle image. Now, the composite shape is nearly a square with side length $x+\frac{b}{2}$; however, there is a corner missing, which has dimensions according with the sides of the rectangle that measure $\frac{b}{2}$; thus, the area of this missing corner are $(\frac{b}{2})^2$. Thus, the “complete square” when we add this little corner to the whole would have the summed areas of the original square ($x^2$), the original rectangle ($bx$), and the additional little square ($(\frac{b}{2})^2$):
$$\text{Area}=x^2+bx+(\frac{b}{2})^2$$
Thus, given some polynomial in the form $p(x)=x^2+bx$, we can “complete the square” in the polynomial by considering that the missing value we need to make this a “perfect square” polynomial is $(\frac{b}{2})^2$. However, we cannot just add something to a function or equation and expect it to remain the same; thus, we need to both add and subtract this value simultaneously to avoid fundamentally altering the function.
$$p(x)=x^2+bx\rightarrow{p(x)=x^2+bx+(\frac{b}{2}^2-(\frac{b}{2}^2}$$
Now let us group that which forms the perfect square polynomial we are attempting to form:
$$p(x)=\bigg(x^2+bx+(\frac{b}{2}^2\bigg)-(\frac{b}{2}^2}$$
And now, let us reduce according to the factorization of a perfect square polynomial:
$$p(x)=\bigg(x^2+bx+(\frac{b}{2}^2\bigg)-(\frac{b}{2}^2}$$
$$p(x)=(x+\frac{b}{2})^2-(\frac{b}{2}^2}$$
This is now in vertex form with a vertex of (-\frac{b}{2}, -(\frac{b}{2})^2)
Now let us consider a more concrete example, using this general form as a guide:
Example: Put the polynomial $x^2+6x$ in vertex form by completing the square.
First, let us recall the form of the perfect square polynomial: $y=x^2+2bx+b^2$
Thus, to make this a perfect square, we should recognize that the coefficient $6=2b$ and so $b=3$. The last term of the perfect square will be $b^2=3^2=9$. And so, $x^2+6x+9$ is the perfect square polynomial we are looking for, but we need to subtract a 9 to compensate for the addition of it:
$$p(x)=x^2+6x+9-9$$
And we group as necessary to form the perfect square:
$$p(x)=\bigg(x^2+6x+9\bigg)-9$$
$$p(x)=(x+3)^2-9$$
Thus, the vertex of this parabola is at $(-3, -9)$
Example: Put the polynomial $x^2-5x+5$ in vertex form by completing the square.
First, do not let the added constant confuse you. We are only considering the $ax^2+bx$ part of the polynomial when completing the square. The added $c=5$ constant will be added into the final result.
Second, let us recall the form of the perfect square polynomial: $y=x^2+2bx+b^2$
Thus, to make this a perfect square, we should recognize that the coefficient $-5=2b$ and so $b=-\frac{5}{2}$. The last term of the perfect square will be $b^2=(-\frac{5}{2})^2=\frac{25}{4}$. And so, $x^2-5x+\frac{25}{4}$ is the perfect square polynomial we are looking for, but we need to subtract a \frac{25}{4} to compensate for the addition of it and bring back the $c$ constant.
$$$x^2-5x+\frac{25}{4}-\frac{25}{4}+5$
And we group as necessary to form the perfect square:
$$p(x)=\bigg(x^2-5x+\frac{25}{4}\bigg)-\frac{25}{4}+5$$
$$p(x)=\bigg(x^2-5x+\frac{25}{4}\bigg)-\frac{25}{4}+\frac{20}{4}$$
$$p(x)=(x-\frac{5}{2})^2-\frac{5}{4}$$
Thus, the vertex of this parabola is at $(\frac{5}{2}, -\frac{5}{4})$
Example: Put the polynomial $3x^2-7x-11$ in vertex form by completing the square.
This third example now gives us a coefficient on $x^2$. Here is what to do in this instance; it is essentially no different except that we should factor out this 3 before completing the square. We also will need to account for this factored out constant when balancing our polynomial with the completion of the square.
Let us recall the form of the perfect square polynomial: $y=x^2+2bx+b^2$
$$y=3x^2-7x-11$$
$$y=3\bigg(x^2-\frac{7}{3}x\bigg)-11$$
Okay, to make this a perfect square, we should recognize that the coefficient $-\frac{7}{3}=2b$ and so $b=-\frac{7}{6}$. The last term of the perfect square will be $b^2=(-\frac{7}{6})^2=\frac{49}{36}$. And so, $x^2-\frac{7}{3}x+\frac{49}{36}$ is the perfect square polynomial we are looking for
Now is when we will need to account for the newly added \frac{49}{36}; however, we also need to account for the fact that there is a $3$ multiplied by the whole perfect square polynomial. Thus, when we subtract \frac{49}{36} to account for its addition, we also need to multiply it by 3.
$$y=3\bigg(x^2-\frac{7}{3}x+\frac{49}{36}\bigg)-11-3(\frac{49}{36})$$
$$y=3(x-\frac{7}{6})^2-\frac{396}{36}-\frac{147}{36}$$
$$y=3(x-\frac{7}{6})^2-\frac{396}{36}-\frac{147}{36}$$
$$y=3(x-\frac{7}{6})^2-\frac{181}{12}$$
Thus, the vertex of this parabola is at $(\frac{7}{6}, -\frac{181}{12})$
Example: Put the polynomial $-2x^2+3x+4$ in vertex form by completing the square.
This last example now gives us a negative coefficient on $x^2$. Here is what to do in this instance; it is essentially no different except that we should factor out both the 2 and -1 before completing the square. We also will need to account for this factored out negative constant when balancing our polynomial with the completion of the square.
Let us recall the form of the perfect square polynomial: $y=x^2+2bx+b^2$
$$y=-2x^2+3x+4$$
$$y=-2\bigg(x^2-\frac{3}{2}x\bigg)+4$$
Okay, to make this a perfect square, we should recognize that the coefficient $-\frac{3}{2}=2b$ and so $b=-\frac{3}{4}$. The last term of the perfect square will be $b^2=(-\frac{3}{4})^2=\frac{9}{16}$. And so, $x^2-\frac{3}{2}x+\frac{9}{16}$ is the perfect square polynomial we are looking for.
Now is when we will need to account for the newly added \frac{9}{16}; however, we also need to account for the fact that there is a $-2$ multiplied by the whole perfect square polynomial. Thus, when we subtract \frac{9}{16} to account for its addition, we also need to multiply it by -2.
$$y=-2\bigg(x^2-\frac{3}{2}x+\frac{9}{16}\bigg)+4-(-2)(\frac{9}{16})$$
$$y=-2(x^2-\frac{3}{4})^2+4+\frac{18}{16}$$
$$y=-2(x^2-\frac{3}{4})^2+\frac{64}{16}+\frac{18}{16}$$
$$y=-2(x^2-\frac{3}{4})^2+\frac{82}{16}$$
$$y=-2(x^2-\frac{3}{4})^2+\frac{41}{8}$$
Thus, the vertex of this inverted parabola (since there is a negative leading coefficient) is at $(\frac{3}{4}, \frac{41}{8})$
Really, if we want the vertex and have the factors, this is easy: we just look for the x-value directly between the x-intercepts and then use this value to find the corresponding y-value. Furthermore, if we have the vertex form, we can find the factors by expanding and simplifying the vertex form into standard form and then factoring it.
So, what’s the use of completing the square to put a second degree polynomial in vertex form if we can find the vertex given the factors of the parabola? Well, it turns out that completing the square has many applications further into mathematics. So, we will return to this concept later. However, there is one very powerful use of completing the square when it comes to quadratics.
What if we cannot find the factors of the parabola? What if the factors are irrational numbers or imaginary numbers, or simply very complex rational numbers. These kind of solutions will be very difficult, if not impossible, to intuit. Is there a general formula that will always guarantee us the factors of the polynomial? Well, yes! However, to get this formula, we will need to complete the square, but not on just any polynomial, but every polynomial all at once. In the next section, we will derive the quadratic formula for the roots of all second degree polynomials by completing the square for the general standard form of a parabola: $ax^2+bx+c$.
5. Quadratic Polynomials: Derivation of and Factoring with the Quadratic Equation
What if we cannot find the factors of a polynomial. We might be able to put it into vertex form, but this does not tell us where the parabola crosses the x-axis. Consider the second example problem from the previous section on completing the square.
$$f(x)=x^2-5x+5$$
We found the vertex form of this parabola through completing the square:
$$f(x)=(x-\frac{5}{2})^2-\frac{5}{4}$$
Thus, the vertex of this parabola is at $(\frac{5}{2}, -\frac{5}{4})$. However, look at where the graph crosses the x-axis: somewhere between 1 and 2 as well as 3 and 4. However, it certainly does not look like 1.5 and 3.5, nor should we assume that it is even if it looked so. What are the factors of this polynomial?
Let’s try using our prior techniques:
$M=5$ and has factors $\pm1$ and $\pm5$. Now, is there any way we can add $\pm1$ and $\pm5$ together such that the equal $N=-5$ and multiply together to equal $M=5$. The answer is no, but in case you are not sure, below are the possibilities:
$$f(x)=x^2-5x+5$$
$(-1)(-5)=5\rightarrow{-1+-5=-6}$
$(1)(5)=5\rightarrow{1+5=6}$
In neither case of multiplying to equal $M$ do the factors also add to equal $N$; thus, our method fails. We cannot factor this polynomial using conventional intuition or the integer factoring technique from previous sections. The solution is actually quite complicated, though the vertex is not so complicated. What might the factors be?
Furthermore, consider this interesting polynomial: $y=x^(2)+x+1$. Seems simple enough, but we encounter the same problem when we use our current factoring techniques. Let’s examine the graph of the function though:
How might we go about factoring such a seemingly non-factorizable polynomial? Well, let’s develop a new and more powerful tool that will allow us to rapidly factor any quadratic polynomial, even if it does not cross the x-axis at all!
Thinking back on our basic algebra techniques, finding the necessary value of $x$ requires us to “isolate the variable with a coefficient of 1. Can we do this with a quadratic in the form $ax^2+bx+c$? Well try as you might, there is not an obvious way to get $x$ by itself using basic algebra techniques. However, consider the vertex form of any parabola:
$$y=a(x-h)^2+k$$
Turn out that we can use basic algebra to solve for x in these cases:
$$y-k=a(x-h)^2+k-k$$
$$\frac{y-k}{a}=\frac{a(x-h)^2}{a}$$
$$\frac{y-k}{a}=(x-h)^2$$
Now we undo the square power on the right by square rooting both sides of the equation.
$$\sqrt{\frac{y-k}{a}]}sqrt{(x-h)^2}$$
$$\pm\sqrt{\frac{y-k}{a}}=x-h$$
$$h\pm\sqrt{\frac{y-k}{a}}=x-h+h$$
$$h\pm\sqrt{\frac{y-k}{a}}=x$$
Ah! So using the vertex form of a parabola, we can isolate the variable $x$ with a coefficient of 1.
So now, let us use this knowledge to find the solutions to $f(x)=x^2-5x+5$. First, we already know the parabola’s vertex form; it is $f(x)=(x-\frac{5}{2})^2-\frac{5}{4}$.
Let do some basic algebra and see if the solutions we get make sense given the graph of the function. We should start by setting $y=0$, since we are looking for our solutions there:
$$0=(x-\frac{5}{2})^2-\frac{5}{4}$$
$$0+\frac{5}{4}=(x-\frac{5}{2})^2-\frac{5}{4}+\frac{5}{4}$$
$$\frac{5}{4}=(x-\frac{5}{2})^2$$
$$\sqrt{\frac{5}{4}}=\sqrt{(x-\frac{5}{2})^2}$$
$$\pm\sqrt{\frac{5}{4}}=x-\frac{5}{2}$$
$$\frac{5}{2}\pm\sqrt{\frac{5}{4}}=x-\frac{5}{2}+\frac{5}{2}$$
$$\frac{5}{2}\pm\sqrt{\frac{5}{4}}=x$$
Thus, the solutions to this parabola (pictured again here) are
$$x=\frac{5}{2}+\sqrt{\frac{5}{4}}=\frac{5}{2}+\frac{\sqrt{5}}{2}=\frac{5+\sqrt{5}}{2}$$
$$x=\frac{5}{2}-\sqrt{\frac{5}{4}}=\frac{5}{2}-\frac{\sqrt{5}}{2}=\frac{5-\sqrt{5}}{2}$$
Or by approximation
$$x\approx3.618$$
$$x\approx1.382$$
Visually, this looks about right!
This is excellent! If we have the vertex form of any parabola, we can isolate the independent variable with a coefficient of 1. However, most parabolas are not given in this form; rather, quadratics are more often seen in standard form (ergo, “standard”). Obviously, given what we have learned, we can put any standard form parabola into vertex form, regardless of the complexity of its solutions. So, can we generalize this method by putting the standard form of all parabolas, $y=ax^2+bx+c$ into vertex form and solving for $x$? Of course we can! First, let’s complete the square on the standard form:
$$y=ax^2+bx+c$$
$$y=\bigg(ax^2+bx\bigg)+c$$
$$y=a\bigg(x^2+\frac{b}{a}x\bigg)+c$$
Okay, to make this a perfect square, we should recognize that the coefficient used to generate our addition little square that completes our larger square (think back to the geometric example) is $\frac{b}{a}=2b_\text{square}$ and so $b_\text{square}=\frac{b}{2a}$. The last term of the perfect square will then be the square made of sides of length $b_\text{square}$ or ${b_\text{square}}^2=(\frac{b}{2a})^2=\frac{b^2}{4a^2}$. And so, the completed square polynomial we are looking for looks like $(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})}$.
Now is when we will need to account for the newly added \frac{b^2}{4a^2}; however, we also need to account for the fact that there is an $a$ multiplied by the whole perfect square polynomial and keep track of the $c$ that was on the outside of the completed square. Thus, when we subtract \frac{b^2}{4a^2} to account for its addition, we also need to multiply it by $a$.
$$a\bigg(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\bigg)+c-(a)(\frac{b^2}{4a^2})$$
Let’s do some simplification:
$$a(x+\frac{b}{2a})^2+c-(\frac{ab^2}{4a^2})$$
$$a(x+\frac{b}{2a})^2+c-(\frac{b^2}{4a})$$
Let’s combine the two constants outside the perfect square polynomial by finding their common denominator:
$$c-\frac{b^2}{4a}$$
$$\frac{4ac}{4a}-\frac{b^2}{4a}$$
$$\frac{4ac-b^2}{4a}$$
So, our standard form quadratic in vertex form looks like:
$$a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$$
From this, we can get the minor result that the vertex of any parabola in standard form $y=ax^2+bx+c$ will be $(-\frac{b}{2a}, \frac{4ac-b^2}{4a})$. But let’s not stop here. Let’s push forward and solve for $x$ when $y=0$
$$a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}=0$$
$$a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}-\frac{4ac-b^2}{4a}=0-\frac{4ac-b^2}{4a}$$
$$a(x+\frac{b}{2a})^2=-\frac{4ac-b^2}{4a}$$
Let’s distribute that negative on the right to make things simpler.
$$a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}$$
$$\frac{a(x+\frac{b}{2a})^2}{a}=\frac{b^2-4ac}{4a(a)}$$
$$x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$$
Now let’s take the square root of both sides:
$$\sqrt{x+\frac{b}{2a})^2}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$$
$$x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{\sqrt{4a^2}}$$
$$x+\frac{b}{2a}-\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$$
$$x=\frac{\pm\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$$
Let’s combine the fractions with their common denominator and invert the order for simplicity’s sake.
$$x=\frac{\pm\sqrt{b^2-4ac}-b}{2a}$$
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Eureka! We have a formula for the solutions to all quadratics in the form $y=ax^2+bx+c$! This is called the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Let’s test this out with three examples: one with known, rational x-intercepts, one with irrational x-intercepts, and one where the graph has no x-intercepts.
Example 1: Rational Solutions with the Quadratic Formula
For this example let’s use the parabola from a previous example on completing the square: $y=x^2+6x$
Graphically, we can easily see that this parabola has solutions and that they are likely rational. We can even take the points $x=-6$ and $x=0$, plug them into the function, and find that the result in both cases is that $y=0$. Does this also work with the quadratic equation? Let’s hope so!
Given the standard form of a parabola $y=ax^2+bx+c$, if $y=x^2+6x$, then $a=1$, $b=6$, and $c=0$. So, we will plug these values into the quadratic equation:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-6\pm\sqrt{6^2-4(1)(0)}}{2(1)}$$
$$x=\frac{-6\pm\sqrt{36-0}}{2}$$
$$x=\frac{-6\pm\sqrt{36}}{2}$$
$$x=\frac{-6\pm6}{2}$$
So,
$$x=\frac{-6+6}{2}=\frac{0}{2}=0$$
$$x=\frac{-6-6}{2}=\frac{-12}{2}=-6$$
Example 2: Irrational Solutions with the Quadratic Formula
For this example let’s use the inverted parabola from a previous example on completing the square: $y=-2x^2+3x+4$
Graphically, we can easily see that this parabola has solutions and that they are possibly irrational or some rational value that is not easily intuited. So, let’s use the quadratic equation to quickly find the solutions.
Given the standard form of a parabola $y=ax^2+bx+c$, if $y=-2x^2+3x+4$, then $a=-2$, $b=3$, and $c=4$. So, we will plug these values into the quadratic equation:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-3\pm\sqrt{3^2-4(-2)(4)}}{2(-2)}$$
$$x=\frac{-3\pm\sqrt{9+32}}{-4}$$
$$x=\frac{-3\pm\sqrt{41}}{-4}$$
So, if we factor out the negative in the top and bottom:
$$x=\frac{3\pm\sqrt{41}}{4}$$
And so,
$$x=\frac{3+\sqrt{41}}{4}\approx{2.35}$$
$$x=\frac{3-\sqrt{41}}{4}\approx{-0.85}$$
These approximate solutions look about right!
Example 3: Imaginary Solutions with the Quadratic Formula
For this example let’s use the parabola that had no x-intercepts from the beginning of this section: $y=x^(2)+x+1$. What might the quadratic equation turn out for us when we know there are no x-intercepts:
Given the standard form of a parabola $y=ax^2+bx+c$, if $y=x^2+x+1$, then $a=1$, $b=1$, and $c=1$. So, we will plug these values into the quadratic equation:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$$
$$x=\frac{-1\pm\sqrt{1-4}}{2}$$
$$x=\frac{-1\pm\sqrt{-3}}{2}$$
And so,
$$x=\frac{-1+\sqrt{-3}}{2}$$
$$x=\frac{-1-\sqrt{-3}}{2}$$
Notice that in both of our solutions, we have the square root of a negative number. For our unit on rooting operations, we learned that the square root of a negative number is not in the “Real Plane on Numbers;” rather, they are Imaginary Numbers. When we combine imaginary numbers like $\sqrt{-1}$ with real numbers, they form “Complex Numbers.” Thus our solutions to this quadratic are in the “Complex Plane of Numbers.” If we wanted to write the solutions as complex numbers, we could also say:
$$x=\frac{-1}{2}\pm\frac{\sqrt{3}}{2}{i}$$
The factored form of our polynomial would then be:
$$y=\bigg(x-(\frac{-1}{2}+\frac{\sqrt{3}}{2}{i})\bigg))\bigg(x-(\frac{-1}{2}-\frac{\sqrt{3}}{2}{i})\bigg)$$
Let’s just work backwards from here and see that we can return the original parabola by distributing these complex numbers:
$$y=\Bigg(x-\bigg(\frac{-1}{2}+\frac{\sqrt{3}}{2}{i}\bigg)\Bigg)\Bigg(x-\bigg(\frac{-1}{2}-\frac{\sqrt{3}}{2}{i}\bigg)\Bigg)$$
$$y=\Bigg(x-\frac{-1+i\sqrt{3}}{2}\Bigg)\Bigg(x-\frac{-1-i\sqrt{3}}{2}\Bigg)$$
$$y=\Bigg(x+\frac{1-i\sqrt{3}}{2}\Bigg)\Bigg(x+\frac{1+i\sqrt{3}}{2}\Bigg)$$
$$y=x^2+x\Bigg(\frac{1+i\sqrt{3}}{2}\Bigg)+x\Bigg(\frac{1-i\sqrt{3}}{2}\Bigg)+\Bigg(\frac{1+i\sqrt{3}}{2}\Bigg)\Bigg(\frac{1-i\sqrt{3}}{2}\Bigg)$$
$$y=x^2+\Bigg(\frac{x+xi\sqrt{3}}{2}\Bigg)+\Bigg(\frac{x-xi\sqrt{3}}{2}\Bigg)+\Bigg(\frac{1+i\sqrt{3}}{2}\Bigg)\Bigg(\frac{1-i\sqrt{3}}{2}\Bigg)$$
$$y=x^2+\Bigg(\frac{(x+xi\sqrt{3})+(x-xi\sqrt{3})}{2}\Bigg)+\Bigg(\frac{(1+i\sqrt{3})(1-i\sqrt{3})}{4}\Bigg)$$
$$y=x^2+\Bigg(\frac{x+x+xi\sqrt{3}-xi\sqrt{3}}{2}\Bigg)+\Bigg(\frac{1-i\sqrt{3}+i\sqrt{3}-(i\sqrt{3})^2}{4}\Bigg)$$
$$y=x^2+\Bigg(\frac{2x+0}{2}\Bigg)+\Bigg(\frac{1-(i\sqrt{3})^2}{4}\Bigg)$$
$$y=x^2+\Bigg(\frac{2x}{2}\Bigg)+\Bigg(\frac{1-(i\sqrt{3})^2}{4}\Bigg)$$
$$y=x^2+x+\Bigg(\frac{1-\bigg((-1)(3)\bigg)}{4}\Bigg)$$
$$y=x^2+x+\Bigg(\frac{1+3}{4}\Bigg)$$
$$y=x^2+x+1$$
Wow! So it seems to be true that a polynomial, when factored, might have imaginary or complex roots!
Example 4: Use the quadratic formula to factor $2x^2 + 12x +3$.
We consider the equation $2x^2 + 12x +3 = 0$ then we note that
$x = \frac{-12 \pm \sqrt{12^2 – 4(2)(3)}}{(2)(2)}$
$x = \frac{-12 \pm \sqrt{120}}{4}$
$x = \frac{-12 \pm 2\sqrt{30}}{4}$
$x = \frac{-6 \pm \sqrt{30}}{2}$
So it seems we should obtain $(x – \frac{-6 + \sqrt{30}}{2})(x – \frac{-6 – \sqrt{30}}{2})$. Let’s multiply these factors out to be sure. We get $(x^2 -\frac{-6 – \sqrt{30}}{2})x – \frac{-6 + \sqrt{30}}{2}x +(\frac{-6 + \sqrt{30}}{2})(\frac{-6 – \sqrt{30}}{2}) = x^2 + 6x + \frac{3}{2}$ What happened? We are off by a factor of 2. The quadratic formula always gives us the factors of the related monic quadratic (quadratic with leading coefficient one). So, we need to rewrite this as:
$2(x – \frac{-6 + \sqrt{30}}{2})(x – \frac{-6 – \sqrt{30}}{2}) = 2x^2 + 12x +3$