9 Factoring

Some factoring forms

$\begin{eqnarray*}&& a^2 - b^2 = (a+ b) (a-b) \\ \\&& a^2 + 2 a b + b^2 = (a+b)^2 \\ \\ && a^2 - 2 a b + b^2 = (a-b)^2 \\ \\ && a^3 + b^3 = (a+b) (a^2 - ab + b^2)\\ \\ && a^3 - b^3 = (a-b) (a^2 + ab + b^2) \end{eqnarray*}$

How to factor a polynomial?

remove a common factor
factor by grouping
use special factoring forms.

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Exercise 1

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$5(x^2+ 4)^4 (2x) (x-2)^4 + (x^2+ 4)^5 (4) (x-2)^3$

$= (x^2+ 4)^4 (x-2)^3 \Big[ 5(2x) (x-2) + (x^2+4) (4)\Big]$

$= (x^2+ 4)^4 (x-2)^3 \Big[ 14x^2 -20 x +16\Big]$ .

Exercise 2

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Set $t= x^2 + 1$ . We have

$(x^2 +1)^2 - 7 (x^2 + 1) + 10 = t^2 - 7 t +10 = ( t - 2) (t-5)$

$= [ x^2 + 1- 2] [ x^2 +1 - 5 ] = (x^2 - 1) (x^2 -4)$

$= (x- 1) (x+1)( x - 2)(x+2)$

We used the identity : $a^2 - b^2 = (a-b) (a+b)$ .

Exercise 3

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We factor the expression by regrouping terms.

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

$x^9 + x^8 + 2x^5 + 2x^4 + x + 1$
$= x^8( x+ 1) + 2x^4( x+ 1) + x + 1$
$=(x+1) (x^8 +2 x^4 + 1 )$
$=(x+1) ( x^4 + 1 )^2$

Thus the common factor is: $(x+1) ( x^4 + 1 )^2$ .

Exercise 4

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We apply the identity: $a^3 + b^3 = (a + b) (a^2 - a b + b^2 )$ .

$x^3 +{27} = x^3 +(3)^3 = (x+{3}) (x^2 - {3} x + {3}^2 ).$

Exercise 5

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We use the method of completing the square. We have

$16 z^2 - 24 z + 9 = (4z)^2 - 2 (4 z) (3) + (3)^2 = (4z-3)^2$

We used the identity : $a^2 -2 a b + b^2 = (a-b)^2$ .

For $8 z^2 + 2 z - 6$ , we can use the same method or the method of determinant.

We have $\Delta = (2)^2- 4(8)(-6)= 4. (49)$ and the roots are

$z_1= (-2-14)/16 =-1, \qquad \qquad z_2 = (-2+14)/16=3/4$

Thus $8 z^2 + 2 z - 6 = 8(z+1) (z-3/4)= 2(z+1)(4z-3).$

Hence, the common factor is : $(4z-3)$ .

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Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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