Rational Roots and Synthetic Division

by Justin Skycak (@justinskycak) on

The rational roots theorem can help us find zeros of polynomials without blindly guessing.

This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Rational Roots and Synthetic Division. In Justin Math: Algebra. https://justinmath.com/rational-roots-and-synthetic-division/


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In the previous post, we learned how to find the remaining zeros of a polynomial if we are given some zeros to start with. But how do we get those initial zeros in the first place, if theyā€™re not given to us and arenā€™t obvious from the equation?

Rational Roots Theorem

The rational roots theorem can help us find some initial zeros without blindly guessing. It states that for a polynomial with integer coefficients, any rational number (i.e. any integer or fraction) that is a root (i.e. zero) of the polynomial can be written as some factor of the constant coefficient, divided by some factor of the leading coefficient.

For example, if the polynomial $p(x)=2x^4+x^3-7x^2-3x+3$ has a rational root, then it is some positive or negative fraction having numerator $1$ or $3$ and denominator $1$ or $2$.

The possible roots are then $\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, or $\pm 3$. We test each of them below.

$\begin{align*} &\mathbf{p\left( \frac{1}{2} \right) = 0} \hspace{.5cm} && p\left( -\frac{1}{2} \right) = \frac{11}{4} \\ & p(1)=-4 \hspace{.5cm} && \mathbf{p(-1)=0} \\ & p\left( \frac{3}{2} \right) = -\frac{15}{4} && p\left( -\frac{3}{2} \right) = -\frac{3}{2} \\ & p(3)=120 && p(-3)=84 \end{align*}$



We see that $x=\frac{1}{2}$ and $x=-1$ are indeed zeros of the polynomial.

Therefore, the polynomial can be written as

$\begin{align*} \left( x-\frac{1}{2} \right) (x+1)(2x^2+bx+c) \end{align*}$


for some constants $b$ and $c$, which we can find by expanding and matching up coefficients.

$\begin{align*} 2x^4+x^3-7x^2-3x+3 &= \left( x-\frac{1}{2} \right) (x+1)(2x^2+bx+c) \\ &= 2x^4+(1+b)x^3+\left( c+\frac{b}{2}-1 \right)x^2 + \left( \frac{c}{2} - \frac{b}{2} \right)x - \frac{c}{2} \end{align*}$


We find that $b=0$ and $c=-6$.

The remaining quadratic factor becomes $2x^2-6$, which has zeros $x=\pm \sqrt{3}$.

Thus, the zeros of the polynomial are $-\sqrt{3}$, $-1$, $\frac{1}{2}$, and $\sqrt{3}$.

Synthetic Division

To speed up the process of finding the zeros of a polynomial, we can use synthetic division to test possible zeros and update the polynomialā€™s factored form and rational roots possibilities each time we find a new zero.

Given the polynomial $x^4+3x^3-5x^2-21x-14$, the rational roots possibilities are $\pm 1$, $\pm 2$, $\pm 7$, and $\pm 14$.

To test whether, say, $2$ is a zero, we can start by setting up a synthetic division template which includes $2$ at the far left, followed by the coefficients of the polynomial (in the order that they appear in standard form).

image


We put a $0$ under the first coefficient (in this case, $1$) and add down the column.

Then, we multiply the result by the leftmost number (in this case, $2$) and put it under the next coefficient (in this case, $3$).

We repeat the same process over and over until we finish the final column.

image


The bottom-right number is the remainder when we divide the polynomial by the factor corresponding to the zero being tested. Therefore, if the bottom-right number is $0$, then the top-left number is indeed a zero of the polynomial, because its corresponding factor is indeed a factor of the polynomial.

In this case, though, the bottom-right number is not $0$ but $-36$, so $2$ is NOT a zero of the polynomial.

However, when we repeat synthetic division with $-2$, the bottom-right number comes out to $0$ and we conclude that $-2$ is a zero of the polynomial.

image


Then $x+2$ is a factor of the polynomial, and the bottom row gives us the coefficients in the sub-polynomial that multiplies to yield the original polynomial.

$\begin{align*} x^4+3x^3-5x^2-21x-14 &= (x+2)(1x^3+1x^2-7x-7) \\ &= (x+2)(x^3+x^2-7x-7) \end{align*}$


The next factor will come from $x^3+x^2-7x-7$, so the rational roots possibilities are just $\pm 1$ and $\pm 7$.

We use synthetic division to test whether $x=1$ is a zero of $x^3+x^2-7x-7$.

image


Since the bottom-right number is $-12$ rather than $0$, we see that $x=1$ is not a zero of $x^3+x^2-7x-7$. However, $x=-1$ is!

image


Using the bottom row as coefficients, we update the factored form of our polynomial.

$\begin{align*} x^4+3x^3-5x^2-21x-14 &= (x+2)(x^3+x^2-7x-7) \\ &= (x+2)(x+1)(1x^2+0x-7) \\ &= (x+2)(x+1)(x^2-7) \end{align*}$


Now that weā€™re down to a quadratic, we can solve it directly.

$\begin{align*} x^2-7 &= 0 \\ x^2 &= 7 \\ x &= \pm \sqrt{7} \end{align*}$


Thus, the zeros of the polynomial are $-2$, $-1$, $\sqrt{7}$, and $-\sqrt{7}$, and the factored form of the polynomial is

$\begin{align*} (x+2)(x+1)(x+\sqrt{7})(x-\sqrt{7}). \end{align*}$


Final Remarks

In this example, the polynomial factored fully into linear factors. However, if the last factor were $x^2+7$, which does not have any zeros, we would leave it in quadratic form. The zeros of the polynomial would be just $-2$ and $-1$, and the fully factored form of the polynomial would be $(x+2)(x+1)(x^2+7)$.

One last thing about synthetic division: be sure to include ALL coefficients of the original polynomial in the top row of the synthetic division setup, even if they are $0$. For example, the polynomial $3x^4+2x$ is really $3x^4+0x^3+0x^2+2x+0$, so the top row in the synthetic division setup should read $\hspace{.5cm} 3 \hspace{.5cm} 0 \hspace{.5cm} 0 \hspace{.5cm} 2 \hspace{.5cm} 0$.

Exercises

For each polynomial, find all the zeros and write the polynomial in factored form. (You can view the solution by clicking on the problem.)

$1) \hspace{.5cm} 3x^3+18x^2+33x+18$
Solution:
$3(x+1)(x+2)(x+3)$
$\text{zeros: } -1, -2, -3$

$2) \hspace{.5cm} 2x^3-5x^2-4x+3$
Solution:
$(2x-1)(x+1)(x-3)$
$\text{zeros: } \frac{1}{2}, -1, 3$

$3) \hspace{.5cm} x^4-4x^3+3x^2+4x-4$
Solution:
$(x+1)(x-1)(x-2)^2$
$\text{zeros: } \pm 1, 2$

$4) \hspace{.5cm} x^4-3x^3+5x^2-9x+6$
Solution:
$(x^2+3)(x-2)(x-2)$
$\text{zeros: } 1, 2$

$5) \hspace{.5cm} x^4-2x^3-x^2+4x-2$
Solution:
$(x-1)^2(x+\sqrt{2})(x-\sqrt{2})$
$\text{zeros: } 1, \pm \sqrt{2}$

$6) \hspace{.5cm} 2x^4+7x^3+6x^2-x-2$
Solution:
$(2x-1)(x+2)(x+1)^2$
$\text{zeros: } \frac{1}{2}, -2, -1$

$7) \hspace{.5cm} 2x^4-8x^3+10x^2-16x+12$
Solution:
$2(x-1)(x-3)(x^2+2)$
$\text{zeros: } 1, 3$

$8) \hspace{.5cm} 21x^5+16x^4-74x^3-61x^2-40x-12$
Solution:
$(x+2)(x-2)(7x+3)(3x^2+x+1)$
$\text{zeros: } \pm 2, -\frac{3}{7}$


This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Rational Roots and Synthetic Division. In Justin Math: Algebra. https://justinmath.com/rational-roots-and-synthetic-division/


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