Factor Theorem

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Question 1 of 2

Factorise

$P(x)=x^3+x^2-14x-24$

1. $(x-3)(x+1)(x-2)$
2. $(x+3)(x-4)(x-1)$
3. $(x+3)(x+2)(x-4)$
4. $x(x+3)(x-4)$

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Long Division

Use long division when a polynomial is divided by a binomial

factor

$(P(x))/(x-a)$ only if

$P(a)=0$

Find the first factor using Factor Theorem and trial and error.

It is easier to try out numbers that divide to the constant term which is

$-24$ .

$24$ can be divided by:

$+-$ $1,2,3,4,6,8,12$

Trying out $a=1$ :

$P(x)$	$=$	$x^3+x^2-14x-24$
$P(a)$	$=$	$a^3+a^2-14a-24$	Replace $x$ with $a$
$P(1)$	$=$	$1^3+1^2-14(1)-24$	Substitute $a=1$
	$=$	$1+1-14-24$
	$=$	$-36$

Since

$P(1)≠0$ ,

$(x-1)$ is not a factor of the polynomial

Trying out $a=2$ :

$P(x)$	$=$	$x^3+x^2-14x-24$
$P(a)$	$=$	$a^3+a^2-14a-24$	Replace $x$ with $a$
$P(2)$	$=$	$2^3+2^2-14(2)-24$	Substitute $a=2$
	$=$	$8+4-28-24$
	$=$	$-40$

Since

$P(2)≠0$ ,

$(x-2)$ is not a factor of the polynomial

Trying out $a=-3$ :

$P(x)$	$=$	$x^3+x^2-14x-24$
$P(a)$	$=$	$a^3+a^2-14a-24$	Replace $x$ with $a$
$P(-3)$	$=$	$(-3)^3+(-3)^2-14(-3)-24$	Substitute $a=-3$
	$=$	$-27+9+42-24$
	$=$	$0$

Since

$P(-3)=0$ , $(x+3)$ is the first factor of the polynomial

Find the second factor using Long Division.

Substitute components into the formula

$\mathsf{P}$ (Polynomial)	$=$	$x^3+x^2-14x-24$
$\mathsf{Divisor}$	$=$	$x+3$

$=$

Next, solve for each term of the quotient

First term of the quotient:

Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial

$x^3dividex$

$=$

$x^2$

Multiply $x^2$ to the divisor. Place this under the Polynomial

$x^2$

$(x+3)$

$=$

$x^3+3x^2$

Subtract $x^3+3x^2$ and write the difference one line below

Drop down

$-14x$ and repeat the process to get the second term of the quotient

Second term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$-2x^2dividex$

$=$

$-2x$

Multiply $-2x$ to the divisor. Place this one line below

$-2x$

$(x+3)$

$=$

$-2x^2-6x$

Subtract $-2x^2-6x$ and write the difference one line below

Drop down

$-24$ and repeat the process to get the third term of the quotient

Third term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$-8xdividex$

$=$

$-8$

Multiply $-8$ to the divisor. Place this under the Polynomial

$-8$

$(x+3)$

$=$

$-8x-24$

Subtract $-8x-24$ and write the difference one line below

Since

$r=0$ and cannot be divided anymore, the quotient is

$x^2-2x-8$

So far, we have factored the polynomial as $(x+3)$

$(x^2-2x-8)$

Factor out the polynomial further by applying cross method

$(x+3)(x^2-2x-8)$

$(x+3)(x+2)(x-4)$

Question 2 of 2

Factorise

$P(x)=2x^3-7x^2-10x+24$

1. $(x-1)(2x-3)(x+2)$
2. $(x+2)(2x-3)(x-4)$
3. $(x+2)(x-3)(x+4)$
4. $x(2x+3)(x-4)$

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Long Division

Use long division when a polynomial is divided by a binomial

Factor Theorem

factor

$(P(x))/(x-a)$ only if

$P(a)=0$

Find the first factor using Factor Theorem and trial and error.

It is easier to try out numbers that divide to the constant term which is

$24$ .

$24$ can be divided by:

$+-$ $1,2,3,4,6,8,12$

Trying out $a=2$ :

$P(x)$	$=$	$2x^3-7x^2-10x+24$
$P(a)$	$=$	$2a^3-7a^2-10a+24$	Replace $x$ with $a$
$P(2)$	$=$	$2(2)^3-7(2)^2-10(2)+24$	Substitute $a=2$
	$=$	$2(8)-7(4)-20+24$
	$=$	$16-28-20+24$
	$=$	$16-28-20+24$
	$=$	$-8$

Since

$P(2)≠0$ ,

$(x-2)$ is not a factor of the polynomial

Trying out $a=-2$ :

$P(x)$	$=$	$2x^3-7x^2-10x+24$
$P(a)$	$=$	$2a^3-7a^2-10a+24$	Replace $x$ with $a$
$P(-2)$	$=$	$2(-2)^3-7(-2)^2-10(-2)+24$	Substitute $a=-2$
	$=$	$2(-8)-7(4)+20+24$
	$=$	$-16-28+20+24$
	$=$	$0$