# Completely factor a polynomial using the rational root theorem and synthetic division

I am currently seriously confused. My problem, as stated above, is about completely factoring a polynomial. My question is, once you get your possible factors, how do you then simplify it down?

Ill give you an example.

2x^4 - x^3 - 21x^2 - 26x - 8

I know how to get the possible zeros (I think).

If I am doing this correct, I believe they are 1/8, 1/4, 1/2, 1, 2, -1/8, -1/4, -1/2, -1, and -2.

With these numbers, how do I completely factor that polynomial?

If I'm not giving enough information, please tell me. Thanks for any help!

• I see that $-1,-2$ are roots in your example, and thus $x+1,x+2$ divides the example. – abiessu Oct 17 '14 at 4:28

By the Rational Roots Theorem, the possible rational roots of $2x^4 - x^3 - 21x^2 - 26x - 8$ are the factors of $-8$ divided by the factors of $2$. Hence, the possible rational roots are

$$\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}$$

By the Remainder Theorem, the remainder when $f(x) = 2x^4 - x^3 - 21x^2 - 26x - 8$ is divided by $x - r$ is $f(r)$. Thus, if $r$ is a root, $f(r) = 0$.

$$f(1) = 2 - 1 - 21 - 26 - 8 = -54 \neq 0$$

so $1$ is not a root.

$$f(-1) = 2 + 1 - 21 + 26 - 8 = 0$$

Hence, $-1$ is a root, so $x - (-1) = x + 1$ is a factor.

Dividing $2x^4 - x^3 - 21x^2 - 26x - 8$ by $x + 1$ yields $2x^3 - 3x^2 - 18x - 8$.

Thus,

$$f(x) = 2x^4 - x^3 - 21x^2 - 26x - 8 = (x + 1)(2x^3 - 3x^2 - 18x - 8)$$

Let $g(x) = 2x^3 - 3x^2 - 18x - 8$. By the Rational Roots Theorem, the possible rational roots of $2x^3 - 3x^2 - 18x - 8$ are the same as those of $f(x)$. We know $1$ is not a root of $f(x)$, so it cannot be a root of $g(x)$.

$$g(-1) = -2 - 3 + 18 - 8 = 5 \neq 0$$

so $-1$ is not a root of $g(x)$, which means it is not a double root of $f(x)$.

$$g(2) = 16 - 12 - 36 - 8 = -40 \neq 0$$

Hence, $2$ is not a root of $g(x)$, which also means it is not a root of $f(x) = (x + 1)g(x)$.

$$g(-2) = -16 - 12 + 36 - 8 = 0$$

Hence, $-2$ is a root of $g(x)$ and $f(x) = (x + 1)g(x)$. Hence, $x - (-2) = x + 2$ is a factor of $g(x)$. Dividing $g(x) = 2x^3 - 3x^2 - 18x - 8$ by $x + 2$ yields $2x^2 - 7x - 4$. Hence,

$$g(x) = 2x^3 - 3x^2 - 18x - 8 = (x + 2)(2x^2 - 7x - 4)$$

so

\begin{align*} f(x) & = 2x^4 - x^3 - 21x^2 - 26x - 8\\ & = (x + 1)g(x)\\ & = (x + 1)(x + 2)(2x^2 - 7x - 4) \end{align*}

To split the linear term of $2x^2 - 7x - 4$, we must find two numbers with product $2 \cdot -4 = -8$ and sum $-7$. They are $-8$ and $1$. Hence,

\begin{align*} 2x^2 - 7x - 4 & = 2x^2 - 8x + x - 4 && \text{split the linear term}\\ & = 2x(x - 4) + 1(x - 4) && \text{factor by grouping}\\ & = (2x + 1)(x - 4) && \text{extract the common factor} \end{align*}

Hence,

\begin{align*} f(x) & = 2x^4 - x^3 - 21x^2 - 26x - 8\\ & = (x + 1)(2x^3 - 3x^2 - 18x - 8)\\ & = (x + 1)(x + 2)(2x^2 - 7x - 4)\\ & = (x + 1)(x + 2)(2x + 1)(x - 4) \end{align*}

which you can check by multiplying the factors.

• Thank you so much. I've been struggling on this question for a while... I appreciate the help! Sorry I couldn't reply earlier. – MurphyTheTurtle Oct 21 '14 at 1:16

In your case should be $2(x-a)(x-b)(x-c)(x-d)$, where $a,b,c,d$ are roots. However it seems strange that a degree 4 polynomial has more than 4 roots...

• How might I go about finding the values of a, b, c, and d? Those numbers do not represent the actual roots, but they are just possible roots. – MurphyTheTurtle Oct 17 '14 at 4:21
• As far I know, there are no generic formulas for 4th degree polynomials. Try to write as product (ax^2+bx+c)(dx^2+ex+f), or (ax+b)(cx^3+dx^2+ex+f), and see if you can find some $a,b,c,d,e,f$ which work. – Milly Oct 17 '14 at 4:23
• If you know all candidates for roots, just test them and see which are really roots. Then use the hint in my answer. – Milly Oct 17 '14 at 4:35