# Synthetic division of polynomials Calculator

## Get detailed solutions to your math problems with our Synthetic division of polynomials step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here!

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### Difficult Problems

1

Solved example of synthetic division of polynomials

$factor\left(x^4+x^3-6x^2-4x+8\right)$
2

We can factor the polynomial $x^4+x^3-6x^2-4x+8$ using synthetic division (Ruffini's rule). We search for a root in the factors of the constant term $8$ and we found that $1$ is a root of the polynomial

$1^4+1^3-6\cdot 1^2-4\cdot 1+8=0$
3

Let's divide the polynomial by $x-1$ using synthetic division. First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

$\left|\begin{array}{c}1 & 1 & -6 & -4 & 8 \\ & 1 & 2 & -4 & -8 \\ 1 & 2 & -4 & -8 & 0\end{array}\right|1$
4

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $x-1$

$\left(x^{3}+2x^{2}-4x-8\right)\left(x-1\right)$
5

We can factor the polynomial $\left(x^{3}+2x^{2}-4x-8\right)$ using synthetic division (Ruffini's rule). We search for a root in the factors of the constant term $-8$ and we found that $2$ is a root of the polynomial

$2^{3}+2\cdot 2^{2}-4\cdot 2-8=0$
6

Let's divide the polynomial by $x-2$ using synthetic division. First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $2$. Add the result to the second coefficient and then multiply this by $2$ and so on

$\left|\begin{array}{c}1 & 2 & -4 & -8 \\ & 2 & 8 & 8 \\ 1 & 4 & 4 & 0\end{array}\right|2$
7

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $x-2$

$\left(x^{2}+4x+4\right)\left(x-2\right)\left(x-1\right)$
8

The trinomial $\left(x^{2}+4x+4\right)$ is perfect square, because it's discriminant is equal to zero

$\Delta=b^2-4ac=4^2-4\left(1\right)\left(4\right) = 0$
9

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^{2}}\:and\:b=\sqrt{4}$
10

Factoring the perfect square trinomial

$\left(x+2\right)^{2}\left(x-2\right)\left(x-1\right)$

$\left(x+2\right)^{2}\left(x-2\right)\left(x-1\right)$