Rational Roots and Synthetic Division
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.
We see that $x=\frac{1}{2}$ and $x=-1$ are indeed zeros of the polynomial.
Therefore, the polynomial can be written as
for some constants $b$ and $c$, which we can find by expanding and matching up coefficients.
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).

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.

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.

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.
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 $x^3+x^2-7x-7$.

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!

Using the bottom row as coefficients, we update the factored form of our polynomial.
Now that weā€™re down to a quadratic, we can solve it directly.
Thus, the zeros of the polynomial are $-2$, $-1$, $\sqrt{7}$, and $-\sqrt{7}$, and the factored form of the polynomial is
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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