因子定理

Just as how we can factor numbers (for example, $12=2^2\cdot 3$), we can also factor polynomials by decomposing it into a product of one or more smaller polynomials. Let's start with an example:

What is the factored form of the quadratic $R(x)=x^2-5x+6$?

$$R(x)=(x-2)(x-3)$$

If you remember, we are breaking apart the middle term to make the polynomial factorable. This is called factoring by grouping.

Now, let's see if we used the same methods!

One way to factor this is to observe that $$x^2-5x+6=x^2-3x-2x+6=x(x-3)-2(x-3)=(x-2)(x-3).$$ Let's think about what would happen if we solve the equation $R(x)=0$. We know by the Quadratic Formula that $R(2)=R(3)=0$, or that $x=2$ and $x=3$ are roots of $R(x)$.

Aha! Perhaps it is a coincidence that the numbers $2$ and $3$ also appear in the factored form of $R(x)$. Thankfully it isn't, and this is an example of the Factor Theorem, which states the following (which is proven at the end of this section):

Factor Theorem: If $P(x)$ is a polynomial, then $P(a)=0$ for some value of $a$ if and only if the term $(x-a)$ is a factor of $P(x)$. (Equivalently, $P(x)=(x-a)Q(x)$ for some other polynomial $Q(x)$.)

So, since $R(2)=R(3)=0$, by the Factor Theorem we must have $R(x)=(x-2)(x-3)T(x)$ for some other polynomial $T(x)$. But since the degree of $R(x)$ and $(x-2)(x-3)$ are both $2$, $T(x)$ must be a constant. This yields $$R(x)=x^2-5x+6=c(x-2)(x-3)$$ for some constant value $c$. Comparing the coefficient of $x^2$ on both sides, it follows that $c=1$ and $$R(x)=x^2-5x+6=(x-2)(x-3)$$ as desired.

Now you try!

Now let's look at another method of factoring, called grouping. Consider the polynomial $P(x)=x^3-3x^2-4x+12$. Let's look at the terms two at a time: $$P(x)=(x^3-3x^2)+(-4x+12).$$ Within each group, we can factor out terms that are common to both elements: $$P(x)=x^2(x-3)-4(x-3).$$ Now, we see that both of the new terms share the factor $(x-3)$. Factoring once again, we see that $$P(x)=(x-3)(x^2-4).$$ This is the general method of factoring by grouping.

Now you try!

Let's now look back at the previous example $$P(x)=x^3-3x^2-4x+12=(x-3)(x^2-4).$$ Let's now try to factor the second term: $R(x)=x^2-4$.

By setting $R(x)=0$, we can quickly see that $R(2)=R(-2)=0$. So, using the Factor Theorem from earlier, we can determine that $$R(x)=(x+2)(x-2).$$ In fact, this nice factorization can be generalized to anything of the form $x^2-y^2$. To do so, let's let $f(x)=x^2-y^2$. Setting $$f(x)=x^2-y^2=0,$$ we can see that $f(y)=f(-y)=0$. From there, it follows by the Factor Theorem that $$f(x)=x^2-y^2=(x+y)(x-y).$$ This factorization is called the difference of squares.

Now, let's think about the polynomial $g(x)=x^3-y^3$. By setting $g(x)=0$, we again see that $g(y)=0$. So, by the Factor Theorem, $$g(x)=(x-y)h(x),$$ for some polynomial $h(x)$.

Let's think about how we can solve for $h(x)$. To do so, remember that for any polynomials $P(x)$ and $Q(x)$, $\deg{P(x)Q(x)}=\deg{P(x)}+\deg{Q(x)}$. Since $\deg{g(x)}=3$ and $\deg{(x-y)}=1$, we must have $\deg{h(x)}=2$. So, $h(x)$ is a quadratic, and we can write $$h(x)=ax^2+bx+c,$$ for some constants $a,b$, and $c$. Then it follows that $$g(x)=x^3-y^3=(x-y)(ax^2+bx+c)=ax^3+(b-ay)x^2+(c-by)x-cy.$$ Comparing the matching coefficients in both sides of the equation, we get the following relationships: $$a=1.$$ $$b-ay=0.$$ $$c-by=0.$$ $$-cy=-y^3.$$ Since $a=1$, plugging in that value into the second equation yields $b=y$. Similarly, $c=y^2$, and so we arrive at our final factorization: $$g(x)=x^3-y^3=(x-y)(x^2+xy+y^2).$$

(Side note: this general method of letting $h(x)$ be an arbitrary polynomial and comparing coefficients---this method is often called "the method of undetermined coefficients"---can be used to factor many other polynomials.)

We have $x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-2}+y^{n-1})$. This is often called a "difference of powers" factorization. In fact, this is how you prove the Factor Theorem above in the first place! :)

The Factor Theorem states that the expression $x-a$, where $a$ is a constant, is a factor of a polynomial $p(x)$ if and only if $p(a)=0$.

How can we begin to prove that if $x-a$ is a factor of $p(x)$, then $p(a)=0$?

Let $p(x)=q(x)(x-a)$.

Simply plug in $a$ to get $p(a)=q(a)(a-a)=0$, as desired. Now we need to prove the other direction since this is an "if and only if" statement.

Plugging in $a$, $p(a)=q(a)(a-a)+r(a)=r(a)$. We know that $p(a)=0$, so $r(a)$ must be $0$ as well. However, since $(x-a)$ is a linear expression, we know that the remainder will be a constant implying that $r(x)=0$ for all $x$. Now let's test what you've learned!

(Video) Factoring Polynomials Using Factor Theorem

by MrsALovesMath

This video by MrsALovesMath walks through an example of using factor theorem to factor polynomials.

Summary

We've previously seen how to factor quadratics with a degree of $2$, but now we will look at factoring polynomials with a degree greater than $2$. The steps to factor this type of polynomial is a little different.

Let's walk through an example, using the Factor Theorem to factor: $4x^4 + 6x^3 -6x^2 -4x$

First, we try to take out the greatest common factor. We can pull out $2x$ from every term.

$$2x(2x^3 +3x^2 -3x -2)$$

$2x$ will be a factor in our final answer, but we've reduced our original polynomial to: $2x^3 +3x^2 -3x -2$, which we will call $P(x)$. $P(x)$ is a three-degree polynomial.

To factor $P(x)$, we will use the Factor Theorem. The Factor Theorem says that if $(x-a)$ is a factor of the polynomial, then the remainder will be equal to $0$. We can find the remainder of a polynomial using the Remainder Theorem. So, if the divisor is $x-a$, then $P(a)=R=0$.

The last constant term of the polynomial is going to be key in this step. The factors of the constant term will be our guesses. We'll start subbing in these values until we get a remainder of $0$.

$$\begin{align} P(1) &= 2(1)^3 + 3(1)^2 -3(1) -2 \\ &= 2 + 3 -3 -2 \\ &= 0 \end{align}$$

In our first guess, we found a remainder of $0$! That means that $x-1$ is one of our factors. Our next step is to divide the polynomial.

$$\frac{2x^3 + 3x^2 -3x -2}{x-1}$$

Long division tells us that the quotient for this expression is: $2x^2 +5x +2$. Could we factor this quotient? We can! We can decompose this quadratic to: $(x+2)(2x+1)$.

So, we have found that:

$$4x^4 + 6x^3 -6x^2 -4x = (2x)(x-1)(x+2)(2x+1)$$

Our original polynomial is on the left side of the equation.

Factor Theorem

An extension of the Remainder Theorem is the Factor Theorem!

The Factor Theorem tells us that, for any polynomial $f(x)$,:

If $f(a)=0$, then $(x-a)$ is a factor of $f(x)$.

We can also flip it and say the opposite! If $(x-a)$ is a factor of $f(x)$, then $f(a)=0$.

Looking Further

In the next lesson, we will learn about the sums and differences of cubes. Descartes’ Rule of Signs helps predict the number of positive and negative zeros, which can then be tested using the factor theorem.