Solved example of synthetic division of polynomials
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 $8$
Next, list all divisors of the leading coefficient $a_n$, which equals $1$
The possible roots $\pm\frac{p}{q}$ of the polynomial $x^4+x^3-6x^2-4x+8$ will then be
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
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
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 $\left(x-2\right)$
For easier handling, reorder the terms of the polynomial $\left(3x^{2}+x^{3}-4\right)$ from highest to lowest degree
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 $-4$
Next, list all divisors of the leading coefficient $a_n$, which equals $1$
The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{3}+3x^{2}-4\right)$ will then be
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
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
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 $\left(x+2\right)$
Factor the trinomial $\left(x^{2}+x-2\right)$ finding two numbers that multiply to form $-2$ and added form $1$
Thus
When multiplying two powers that have the same base ($x+2$), you can add the exponents
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