Question

i
factor each of the following expressions.

• $9x^{2}-1$
• $x^{2}-7x + 10$
• $x^{2}+12x + 36$
• $2x^{2}-24x$
• $5x^{2}-x - 18$

ii
factor completely.

• $4x^{2}+52x + 168$
• $3x^{3}-48x$
• $x^{4}-81$
• $x^{4}-23x^{2}-50$
• $2x^{4}-32y^{8}$

Explanation:

Step1: Factor $9x^2-1$ (difference of squares)

$9x^2-1=(3x)^2-(1)^2=(3x-1)(3x+1)$

Step2: Factor $x^2-7x+10$ (trinomial)

Find two numbers: $-2,-5$; product=10, sum=-7.
$x^2-7x+10=(x-2)(x-5)$

Step3: Factor $x^2+12x+36$ (perfect square)

$x^2+12x+36=(x+6)^2$

Step4: Factor $2x^2-24x$ (GCF first)

GCF is $2x$: $2x^2-24x=2x(x-12)$

Step5: Factor $5x^2-x-18$ (trinomial)

Find two numbers: $9,-10$; product $5*(-18)=-90$, sum=-1.
Split middle term: $5x^2+9x-10x-18=x(5x+9)-2(5x+9)=(5x+9)(x-2)$

Step6: Factor $4x^2+52x+168$ (GCF first)

GCF is 4: $4(x^2+13x+42)$. Factor trinomial: $4(x+6)(x+7)$

Step7: Factor $3x^3-48x$ (GCF then difference of squares)

GCF is $3x$: $3x(x^2-16)=3x(x^2-4^2)=3x(x-4)(x+4)$

Step8: Factor $x^4-81$ (difference of squares twice)

$x^4-81=(x^2)^2-9^2=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9)$

Step9: Factor $x^4-23x^2-50$ (substitute $u=x^2$)

Let $u=x^2$: $u^2-23u-50=(u-25)(u+2)=(x^2-25)(x^2+2)=(x-5)(x+5)(x^2+2)$

Step10: Factor $2x^4-32y^8$ (GCF then difference of squares)

GCF is 2: $2(x^4-16y^8)=2((x^2)^2-(4y^4)^2)=2(x^2-4y^4)(x^2+4y^4)=2((x)^2-(2y^2)^2)(x^2+4y^4)=2(x-2y^2)(x+2y^2)(x^2+4y^4)$

Answer:

1. $9x^2 - 1 = (3x-1)(3x+1)$
2. $x^2 - 7x + 10 = (x-2)(x-5)$
3. $x^2 + 12x + 36 = (x+6)^2$
4. $2x^2 - 24x = 2x(x-12)$
5. $5x^2 - x - 18 = (5x+9)(x-2)$
6. $4x^2 + 52x + 168 = 4(x+6)(x+7)$
7. $3x^3 - 48x = 3x(x-4)(x+4)$
8. $x^4 - 81 = (x-3)(x+3)(x^2+9)$
9. $x^4 - 23x^2 - 50 = (x-5)(x+5)(x^2+2)$
10. $2x^4 - 32y^8 = 2(x-2y^2)(x+2y^2)(x^2+4y^4)$