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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 the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $8$

$1, 2, 4, 8$
3

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

$1$
4

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^4+x^3-6x^2-4x+8$ will then be

$\pm1,\:\pm2,\:\pm4,\:\pm8$
5

Trying all possible roots, we found that $2$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

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

Now, divide the polynomial by the root we found $\left(x-2\right)$ using synthetic division (Ruffini's rule). 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

$\left|\begin{array}{c}1 & 1 & -6 & -4 & 8 \\ & 2 & 6 & 0 & -8 \\ 1 & 3 & 0 & -4 & 0\end{array}\right|2$
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 $\left(x-2\right)$

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

We can factor the polynomial $\left(x^{3}+3x^{2}-4\right)$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $-4$

$1, 2, 4$
9

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

$1$
10

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{3}+3x^{2}-4\right)$ will then be

$\pm1,\:\pm2,\:\pm4$
11

Trying all possible roots, we found that $-2$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

${\left(-2\right)}^{3}+3\cdot {\left(-2\right)}^{2}-4=0$
12

Now, divide the polynomial by the root we found $\left(x+2\right)$ using synthetic division (Ruffini's rule). 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

$\left|\begin{array}{c}1 & 3 & 0 & -4 \\ & -2 & -2 & 4 \\ 1 & 1 & -2 & 0\end{array}\right|-2$
13

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 $\left(x+2\right)$

$\left(x^{2}+x-2\right)\left(x+2\right)\left(x-2\right)$
14

Factor the trinomial $\left(x^{2}+x-2\right)$ finding two numbers that multiply to form $-2$ and added form $1$

$\begin{matrix}\left(-1\right)\left(2\right)=-2\\ \left(-1\right)+\left(2\right)=1\end{matrix}$
15

Thus

$\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+2\right)$
16

When multiplying two powers that have the same base ($x+2$), you can add the exponents

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

Final Answer

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

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