# Factoring Difference of Squares help? 5 problems.?

Can you please do these 5 problems and show me your work? I just am not understanding how to do them.

1) 2p^3 - 162p

2) u^4 - 81

3) 2d^4 - 32f^4

4) 20r^4 - 45n^4

5) 256n^4 - c^4

Relevance
• 8 years ago

1)2p³ - 162p

=2p(p² - 81)

=2p(p² - 9²)

=2p (p + 9) (p - 9)

2)u⁴ - 81

=u⁴ - 3⁴

=(u²)² - (3²)²

=(u² + 3²) (u² - 3²)

=(u² + 9) (u + 3) (u - 3)

3)2d⁴ - 32f⁴

=2(d⁴ - 16f⁴)

=2{ (d²)² - (4f²)² }

=2 { (d² + (4f)² } { d² - (4f)² }

=2 (d² + 4f²) ( d + 4f) (d - 4f)

4)20r⁴ - 45n⁴

=5(4r⁴ - 9n⁴)

=5{ (2r²)² - (3n²)² }

=5 (2r² + 3n²) (2r² - 3n²)

5)256n⁴ - c⁴

=(16n²)² - (c²)²

=(16n² + c²) (16n² - c²)

=(16n² + c²) { (4n)² - (c)² }

=(16n² + c²) (4n + c) (4n - c)

• Colin
Lv 7
8 years ago

Fundamental truth:

a^2 - b^2 = (a - b)*(a + b)

3) 2d^4 - 32f^4

= 2*(d^4 - 16*f^4)

Recognise as difference of squares

= 2*([d^2]^2 - [4*f^2]^2)

= 2*(d^2 -4*f^2)*(d^2 +4*f^2)

First term is difference of squares again

= 2*(d - 2*f)*(d + 2*f)*(d^2 +4*f^2) <<<

4) 20r^4 - 45n^4

Take out a factor of 5...

5) 256n^4 - c^4

= ([16*n^2]^2 - [c^2]^2)

and off you go

• 4 years ago

(a^2)-4 = 0 (a-2) (a+2) = 0 a-2 = 0 or a+2 = 0 a = 2 or a = -2 a million-49c^2 a million-40 9 c^2 = 0 -((7 c-a million) (7 c+a million)) = 0 multiply the two area cases -a million to do away with the - (7 c-a million) (7 c+a million) = 0 7 c-a million = 0 or 7 c+a million = 0 7 c = a million or 7 c = -a million c = a million/7 or c = -a million/7

Lv 7
8 years ago

2p³ - 162p = 2p(p² - 81)

= 2p(p + 9)(p - 9)

:::::

u⁴ - 81 = (u²)² - 9²

= (u² + 9)(u² - 9)