Answer to Question #240907 in Calculus for CCW

Question

Factoring Polynomials Practice:

Use synthetic division to divide polynomials by possible roots

x³ + 4x² + 5x + 2

P= Q=

Answer 1

Since all coefficients are integers, we can apply the rational zeros theorem.

The trailing coefficient (the coefficient of the constant term) is $2.$

Find its factors (with the plus sign and the minus sign): $±1,±2.$

These are the possible values for $p.$

The leading coefficient (the coefficient of the term with the highest degree) is $1.$ Find its factors (with the plus sign and the minus sign): $±1.$

These are the possible values for $q.$

Find all possible values of $\dfrac{p}{q}:\pm\dfrac{1}{1}, \pm\dfrac{2}{1} .$

Simplify and remove the duplicates (if any).

These are the possible rational roots: $\pm1, \pm2.$

Next, check the possible roots: if $a$ is a root of the polynomial $P(x),$ the remainder from the division of $P(x)$ by $x−a$ should equal $0.$

Check $1:$ divide $x^3 + 4x^2 + 5x + 2$ by $x-1.$

$P(1)=1+4+5+2=12;$  thus, the remainder is $12.$

Check $-1:$ divide $x^3 + 4x^2 + 5x + 2$ by $x+1.$

$P(-1)=-1+4-5+2=0;$  thus, the remainder is $0.$

Hence, $−1$ is a root.

Check $2:$ divide $x^3 + 4x^2 + 5x + 2$ by $x-2.$

$P(2)=8+16+10+2=36;$  thus, the remainder is $36.$

Check $-2:$ divide $x^3 + 4x^2 + 5x + 2$ by $x+2.$

$P(-2)=-8+16-10+2=0;$  thus, the remainder is $0.$

Hence, $−2$ is a root.

$\def\arraystretch{1.5} \begin{array}{c:c} & x^3 & x^2 & x^1 & x^0 \\ -1 & 1 & 4 & 5 & 2 \\ & & -1 & -3 & -2\\\hline & 1 & 3 & 2 & 0 \\ \end{array}$

$x^3+4x^2+5x+2=(x+1)(x^2+3x+2)$

$\def\arraystretch{1.5} \begin{array}{c:c} & x^2 & x^1 & x^0 \\ -2 & 1 & 3 & 2 \\ & & -2 & -2 \\\hline & 1 & 1 & 0 \\ \end{array}$

$x^2+3x+2=(x+2)(x+1)$

Then

$x^3+4x^2+5x+2=(x+1)^2(x+2)$

Roots are: $-2$ of multiplicity $1, -1$ of multiplicity $2.$