Factoring Polynomials using the Factor Theorem Alternative Format

Lesson Part 1

In This Module

• We will discuss how the remainder theorem, factor theorem, and polynomial division can be used to factor polynomials.

Factoring Polynomials

The ability to factor a polynomial, for example \(2x^2+7x-15=(2x-3)(x+5)\), is essential to graphing polynomial functions and solving polynomial equations.

When a polynomial is in factored form, the zeros of the function, or the roots of the equation, are easily identifiable

You have previously developed skills to factor quadratic polynomials, identifying linear factors that produce rational roots or zeros, as shown below.

\[ \begin{align*} f(x) &= 2x^2+7x-15 \\ 0 &= 2x^2+7x-15 \\ 0 &=(2x-3)(x+5) \end{align*} \]\[ \begin{align*} 2x-3=0 & \text{ or } x+5=0 \\ x=\dfrac{3}{2} & \text{ or } x=-5 \end{align*} \]

Before we begin factoring higher-degree polynomials, we will clarify what we mean when we say factor. Let's pose a question.

Can \(x^2-3\) be factored?

Using traditional factoring techniques, we could conclude that it cannot be factored.

However, \(x^2-3\) can be expressed as \((x+\sqrt{3})(x-\sqrt{3})\) and each binomial is a linear factor of \(x^2-3\).

For our purposes, when factoring polynomials with linear factors, we will consider only linear factors of the form \(mx-n\), where \(\dfrac{m}{n}\) is a rational number.

For example, \(3x^2-11x-4\) is factorable since \(3x^2-11x-4=(3x+1)(x-4)\).

But, \(x^2-2x-1\) will be classified as non-factorable since the roots of the equation \(x^2-2x-1=0\) are irrational.

Factoring Polynomials Using the Factor Theorem

Example 1

Factor \(x^3-4x^2-3x+18\).

We will employ the factor theorem as we begin this process. Recall:

Factor Theorem

For a given polynomial, \(P(x)\), \((x-n)\) is a factor of \(P(x)\) if and only if \(P(n)=0\). This means that if \(P(n)=0\), then \((x-n)\) is a factor of \(P(x)\). Conversely, if \((x-n)\) is a factor of \(P(x)\), then \(P(n)=0\). This can extend to include any linear factor. In other words, \((mx-n)\) is a factor of \(P(x)\) if and only if \(P\left(\frac{n}{m}\right)=0\).

Solution

Let \( P(x) = x^3 - 4x^2 - 3x + 18 \).

Our first step in factoring this polynomial is to find a value, \( x = n \), such that \( P(n) = 0 \). We can randomly try different values for \(x\), hoping to hit on one that works.

\[ \begin{align*} P(1) &= \ ? \\ P(-1) &= \ ? \\ P(2) &= \ ? \end{align*} \]

But there is actually another theorem that will help us narrow down the choice of values that we test.

The rational root theorem stipulates a restriction on the rational roots of a polynomial equation with integer coefficients.

This theorem will provide us with a list of test values for \(x\) that can be used with the factor theorem to find the first factor of the polynomial.

Rational Root Theorem

Given a polynomial equation \( P(x) = 0 \) with integer coefficients, the rational roots of the equation are of the form \( \frac{q}{r} \), where \( q \) is a factor of the constant term of \( P(x) \) and \( r \) is a factor of the leading coefficient (the coefficient of the term of the highest degree).

This means that our test values must satisfy this condition.

Remember: a rational number is any number that can be written in the form \( \frac{a}{b} \), where \(a\) and \(b\) are integers and \(b \neq 0\). Therefore, the rational root theorem encompasses integer roots.

Let's return to the problem and use these theorems to factor this cubic polynomial.

Solution

Let \( P(x) = x^3 - 4x^2 - 3x + 18 \).

• We will always look for a common factor first: this will help to simplify the process. There is no common factor in this example.
• Using the rational root theorem, we create a list of all possible test values for later use with the factor theorem.

We list all combinations of the factors of the constant term, \(18\), divided by the factors of the leading coefficient, \(1\).

For the rational root \( \frac{q}{r} \), \( q \) must divide \(18\), and thus must come from the set \( \{\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \} \).

Similarly, \( r \) must divide into 1, giving two possibilities, \( \{\pm 1 \} \).

\[\tfrac{q}{r} \in \left\{\pm \tfrac{1}{1}, \pm \tfrac{2}{1}, \pm \tfrac{3}{1}, \pm \tfrac{6}{1}, \pm \tfrac{9}{1}, \pm \tfrac{18}{1}\right\}\]

This list, of course, can be simplified to just the factors of \(18\), since the leading coefficient was \(1\).

\[\tfrac{q}{r} \in \{\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \}\]

• Using the factor theorem, we look for a value, \(x=n\), from the test values such that \(P(n)=0\).

You may want to consider the coefficients of the terms of the polynomial and see if you can cut the list down.

Sometimes, if the value of the coefficients or constant term is large, you may want to start with a larger test value. Use some deductive reasoning here.

In this example, we have a mixture of positive and negative coefficients, so there is not necessarily a restriction on the sign of the test value.

With a quick look, we can see that \(\pm 1\) will not work since the constant value is significantly larger than the coefficients of the first three terms. So we will start with \(2\) and \(-2\).

• We use \((x +2)\) as the first factor of the polynomial and we can determine the corresponding factor by long division (or synthetic division).

			\(x^2\)	\(-\)	\(6x\)	\(+\)	\(9\)
\(x+2\)	\(x^3\)	\(-\)	\(4x^2\)	\(-\)	\(3x\)	\(+\)	\(18\)
	\(x^3\)	\(+\)	\(2x^2\)
		\(-\)	\(6x^2\)	\(-\)	\(3x\)
		\(-\)	\(6x^2\)	\(-\)	\(12x\)
					\(9x\)	\(+\)	\(18\)
					\(9x\)	\(+\)	\(18\)
							\(0\)

Using long division, the quotient \(x^2 - 6x + 9\) is our quadratic factor since the remainder is \(0\).

We now have \(x^3-4x^2-3x+18=(x+2)(x^2-6x+9)\)

We can continue by factoring the quadratic. The fully factored form of the cubic would be

\[ \begin{align*} x^3 - 4x^2 - 3x + 18 &= (x + 2)(x - 3)(x - 3) \\ &= (x + 2)(x - 3)^2 \end{align*} \]

I have used long division to find the quadratic factor. You may wish to use synthetic division instead to obtain that corresponding factor.

Check Your Understanding A, B, and C

These questions are not included in the Alternative Format, but can be accessed in the Review section of the side navigation.

Lesson Part 2

Have and Need

Let's look at an alternate approach to both long division or synthetic division. I call this method “have and need.” We work backwards to expanding. Let's consider the same question.

Example 1 Revisited

Factor \(x^3-4x^2-3x+18\).

Solution

Note: The term \(\blacksquare\) represents an unknown coefficient. These coefficients will be determined by comparing like terms in the equations.

\[x^3 - 4x^2 - 3x + 18 = (x+2)(\blacksquare x^2 + \blacksquare x + \blacksquare)\]

We have our first factor, \((x + 2)\), and know that the corresponding factor of this cubic is quadratic since a linear multiplied by a quadratic is cubic.

If we now expand out what we have using the two known terms from each bracket, we have four of the six terms that would come from the full expansion of the binomial times the trinomial.

\[(x+2)(x^2+9)=x^3+2x^2+9x+18\]

Here, we have:

\[x^3+2x^2+9x+18\]

Notice that we have the first and last term of the polynomial that we are factoring.

When the missing middle term of the quadratic factor is multiplied by the linear factor, it will produce two more terms.

When these two terms are added to the four we have already, the result must be the polynomial we have set out to factor.

We determine what terms we need to add to the four terms we have to create the original polynomial.

These terms can be found by subtracting the terms we have from the polynomial we are factoring.

\[ \underbrace{(x^3\color{NavyBlue}-4x^2\color{BrickRed}-3x\color{black}+18)}_{\text{Original}}-\underbrace{(x^3\color{NavyBlue}+2x^2\color{BrickRed}+9x\color{black}+18)}_{\text{Have}}=\underbrace{\color{NavyBlue}-6x^2\color{BrickRed}-12x}_{\text{Need}} \]

We have \(2x^2\), so we need \(-6x^2\) to obtain the \(-4x^2\) in the original polynomial.
We have \(9x\), so we need \(-12x\) to obtain the \(-3x\) in the original polynomial.

We need:

\[-6x^2 - 12x\]

Working with the two terms we need, if we now factor \(-6x\) from these two terms, the remaining factor is \((x + 2)\).

\[\color{NavyBlue}-6x\color{BrickRed}(x+2)\]

If this factor is not \((x+2)\) (i.e., if it is not the same as the first factor of the polynomial), an error has been made somewhere in the solution.

\[\color{BrickRed}(x+2)\color{black}(x^2\color{NavyBlue}-6x\color{black}+9)\]

We can then finish off the factoring process by factoring the quadratic.

Hence,

\[ \begin{align*} x^3 - 4x^2 - 3x + 18 &= (x + 2)(x^2 - 6x + 9) \\ &= (x + 2)(x - 3)(x - 3) \end{align*} \]

Examples

Example 2

Factor \(2x^3+3x^2-17x-30\).

Solution

In this example, there is no common factor to the terms of the polynomial. So we start by determining the test values to use with the factor theorem.

We list all possible factors of the constant term, \(-30\): \( q \in \{\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30\} \)

Divided by all possible factors of the leading coefficient, \(2\): \( r \in \{\pm 1, \pm 2\} \)

\[ \tfrac{q}{r} \in \left\{\pm \tfrac{1}{1}, \pm \tfrac{2}{1}, \pm \tfrac{3}{1}, \pm \tfrac{5}{1}, \pm \tfrac{6}{1}, \pm \tfrac{10}{1}, \pm \tfrac{15}{1}, \pm \tfrac{30}{1}, \pm \tfrac{1}{2}, \pm \tfrac{2}{2}, \pm \tfrac{3}{2}, \pm \tfrac{5}{2}, \pm \tfrac{6}{2}, \pm \tfrac{10}{2}, \pm \tfrac{15}{2}, \pm \tfrac{30}{2}\right\} \]

This list can, of course, be simplified.

\[ \tfrac{q}{r} \in \left\{\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}, \pm \tfrac{5}{2}, \pm \tfrac{15}{2} \right\} \]

These test values don't necessarily need to be listed, but you should be aware of the restrictions on the values you consider testing.

This is a long list but with some luck, we will find a value quickly that will produce a zero value for the polynomial. We will try the integers before testing the fractions.

Let \( P(x) = 2x^3 + 3x^2 - 17x - 30 \).

Again, looking at the coefficients, we can see fairly easily that substituting \(\pm 1\) in for \(x\) will not result in a zero value.

We will move on to testing \(x = \pm 2\).

Using the “have and need” approach,

\[2x^3 + 3x^2 - 17x - 30 = (x + 2)(2x^2 + \blacksquare x - 15)\]

The first term must be \(2x^2\) since \(x \times 2x^2\) will give us the first term of the polynomial, \(2x^3\).
The last term must be \(-15\) since \(2 \times -15\) will give us the last term of the polynomial, \(-30\).

We expand out these four terms and we have:

\[2x^3+4x^2-15x-30\]

Now, the middle term of the quadratic will generate two more terms that when added to the first four terms we have will produce the polynomial we are factoring.

We have \(4x^2\). We need \(-x^2\) to obtain \(3x^2\).
We have \(-15x\). We need \(-2x\) to obtain \(-17x\).

\[-x^2-2x\]

By common factoring the \(-x\) from the two terms we need, we obtain the factor \((x + 2)\), which is identical to the linear factor of our polynomial.

\[\color{NavyBlue}-x\color{BrickRed}(x+2)\]

Therefore, the middle term of the quadratic is \(-x\).

\[2x^3 + 3x^2 - 17x - 30 = \color{BrickRed}(x + 2)\color{black}(2x^2\color{NavyBlue}-x\color{black}-15)\]

We continue then by factoring the quadratic as shown.

\[ \begin{align*} 2x^3 + 3x^2 - 17x - 30 &= (x + 2)(2x^2 -x - 15) \\ &= (x + 2)(2x + 5)(x - 3) \end{align*} \]

We see that \( P(3) = 0 \) and \( P\left(-\frac{5}{2}\right) = 0 \), hence \(( x - 3 )\) or \(( 2x + 5 )\) could have been used to start the factoring process.

Check Your Understanding D

This question is not included in the Alternative Format, but can be accessed in the Review section of the side navigation.

Lesson Part 3

Examples

If the polynomial is a fourth degree polynomial or higher, we need to repeat the process over again until the polynomial is factored as fully as possible.

Example 3

Factor the quartic polynomial \(x^4-3x^3-13x^2+3x+12\).

Solution

Again, the first step should be to look for a common factor. There is no common factor here, so we will move on to the rational root theorem to determine the test values and the factor theorem to find the first factor.

\[ \begin{align*} q \in \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\} &\rightarrow \text{all possible factors of the constant term, 12} \\ r \in \{\pm 1\} &\rightarrow \text{all possible factors of the leading coefficient, 1} \\ \tfrac{q}{r} \in \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\} &\rightarrow\text{all possible test values} \end{align*} \]

Looking at the coefficients of the terms of the polynomial, it appears that \(1\) or \(-1\) may work to produce a zero for the polynomial.

Let \( P(x) = x^4 - 3x^3 - 13x^2 + 3x + 12 \).

We use long division to find the cubic factor.

			\(x^3\)	\(-\)	\(2x^2\)	\(-\)	\(15x\)	\(-\)	\(12\)
\(x-1\)	\(x^4\)	\(-\)	\(3x^3\)	\(-\)	\(13x^2\)	\(+\)	\(3x\)	\(+\)	\(12\)
	\(x^4\)	\(-\)	\(x^3\)
		\(-\)	\(2x^3\)	\(-\)	\(13x^2\)
		\(-\)	\(2x^3\)	\(+\)	\(2x^2\)
				\(-\)	\(15x^2\)	\(+\)	\(3x\)
				\(-\)	\(15x^2\)	\(+\)	\(15x\)
						\(-\)	\(12x\)	\(+\)	\(12\)
						\(-\)	\(12x\)	\(+\)	\(12\)
							\(0\)

Alternatively, we could use synthetic division to find the cubic factor.

\(1\)	\(1\)	\(-3\)	\(-13\)	\(3\)	\(12\)
		\(1\)	\(-2\)	\(-15\)	\(-12\)
	\(1\)	\(-2\)	\(-15\)	\(-12\)	\(0\)

There is a way using an extended “have and need” method that is more challenging. You may want to try this on your own.

Hence, \(x^4 - 3x^3 - 13x^2 + 3x + 12 = (x - 1)(x^3 - 2x^2 - 15x - 12)\).

The quotient found by the division process provides us with our second factor, which in this case is cubic.

\[(x-1)(\textcolor{BrickRed}{x^3-2x^2-15x-12})\]

To factor \( x^3 - 2x^2 - 15x - 12 \), we must now repeat this process and use the rational root theorem and the factor theorem.

Let \(g(x)= x^3-2x^2-15x-12\).

Note that substituting \(-1\) into our cubic expression results in \(0\).

The corresponding quadratic factor can be determined using the “have and need” method as shown here or by long or synthetic division.

Identifying the first and last term gives

\[ x^3 - 2x^2 - 15x - 12 = (x + 1)(x^2 + \blacksquare x - 12) \]

So we have \( x^3 + x^2 - 12x - 12 \) and thus we need \( -3x^2 - 3x = -3x(x + 1) \).

Hence,

\[ x^4 - 3x^3 - 13x^2 + 3x + 12 = (x - 1)(x + 1)(x^2 - 3x - 12) \]

Factoring the quadratic expression found by “have and need” would involve factors with irrational terms. For our purposes, we will deem this quadratic as non-factorable.

The quadratic cannot be factored further.

Therefore, \( x^4 - 3x^3 - 13x^2 + 3x + 12 \) in factored form is \( (x - 1)(x + 1)(x^2 - 3x - 12) \).

Example 4

If the cubic polynomial \( ax^3 + bx^2 + c \) with \( a \neq 0\) and \(c \neq 0 \) has a factor of the form \( x^2 + nx + 1 \), show that \( c^2 - a^2 = bc \).

Pause here so you can try to solve this problem on your own.

Solution

If the given factor of this cubic is the quadratic \(x^2 + nx + 1\), then the other factor must be linear.

Hence,

\[ ax^3 + bx^2 + c = (\blacksquare x + \blacksquare)(x^2 + nx + 1) \]

Using the same concepts we used with have and need, we can determine the first and last term of this linear factor.

The first term of the linear factor must be \(ax\) since \(ax(x^2 )=ax^3\).

\[\textcolor{BrickRed}{ax^3} + bx^2 + c = (\textcolor{BrickRed}{ax} + \blacksquare)(\textcolor{BrickRed}{x^2} + nx + 1) \]

The last term of the linear factor must be \(c\) since the product of the last terms of the two factors must be \(c\).

\[\textcolor{BrickRed}{ax^3} + bx^2 + \textcolor{NavyBlue}{c} = (\textcolor{BrickRed}{ax} + \textcolor{NavyBlue}{c})(\textcolor{BrickRed}{x^2} + nx + \textcolor{NavyBlue}{1}) \]

Hence,

\[ \begin{align*} ax^3 + bx^2 + c &= (ax + c)(x^2 + nx + 1) \\ &= ax^3 + anx^2 + ax + cx^2 + cnx + c \\ &= ax^3 + (an + c)x^2 + (a + cn)x + c \end{align*} \]

By expanding the right-hand side and simplifying, collecting like terms, we can then equate coefficients such that the left side of the equation equals the right side.

\[\textcolor{BrickRed}{a}x^3 + \textcolor{Mulberry}{b}x^2 + \textcolor{Violet}{0}x + \textcolor{NavyBlue}{c} = \textcolor{BrickRed}{a}x^3 + (\textcolor{Mulberry}{an + c})x^2 + (\textcolor{Violet}{a + cn})x + \textcolor{NavyBlue}{c}\]

By equating the coefficients, we have two equations.

\[ \begin{align*} an + c &= b \tag{1} \\ a + cn &= 0 \tag{2} \end{align*} \]

Rearranging equation \((2)\) gives

\[ n = -\frac{a}{c} \text{ since } c \neq 0 \]

Substituting into equation \((1)\),

\[ a\left(-\frac{a}{c}\right) + c = b \]

and multiplying by \( c \) (since \( c \neq 0 \))

\[ \begin{align*} -a^2 + c^2 &= bc \\ c^2 - a^2 &= bc \end{align*} \]

Lesson Part 4

Summary

• To factor a polynomial, \(P(x)\), of degree \(3\) or greater, begin with the factor theorem. Look for a value \(n\), such that \(P(n)=0\).
• To employ the rational root theorem (also called the rational zero theorem), the test values for \(n\) must be of the form \(\frac{q}{r}\), where \(q\) is a factor of the constant term of \(P(x)\) and \(r\) is a factor of the leading coefficient. If \(P\left(\frac{q}{r}\right)=0\), then \(rx-q\) is a factor of \(P(x)\).
• To determine the other factor(s) of \(P(x)\), divide \(rx-q\) into the polynomial using long division (or synthetic division) or use the “have and need” method.
• This process may need to be repeated several times depending on the degree of the polynomial.
• It is possible that no linear factors will be found depending on the nature of the polynomial.