Question

Factor the polynomial 81x^4−16y^4

Answer 1

@primeralph

Answer 2

@ranga

Answer 3

Use the identity: a^2 - b^2 = (a + b)(a - b)

Answer 4

(9-2)(9+2)

Answer 5

81x^4−16y^4 = (9x^2)^2 - (4y^2)^2 = (9x^2 + 4y^2)(9x^2 - 4y^2)

You can further factor (9x^2 - 4y^2) using the exact same method again.

Answer 6

(3x-2y) (3x+2y)??

Answer 7

Yes.

Answer 8

2x^3+16y^3?

Answer 9

new question ^^

Answer 10

Factor out 2 first.
Then use the identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Answer 11

2(x^2+8y^2)

Answer 12

2x^3+16y^3 = 2(x^3 + 8y^3) = 2{ (x)^3 + (2y)^3 } = ?
Use the above identity for the terms inside { }

Answer 13

wont the exponent be 2 instead of 3?

Answer 14

No. You are just factoring out a number 2 first.

Answer 15

can you show me step by step for to get the answer?

Answer 16

2x^3+16y^3 = 2(x^3 + 8y^3) = 2{ (x)^3 + (2y)^3 }.
Use the identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2)
2{ (x)^3 + (2y)^3 } = 2{ (x + 2y)( x^2 - (x)(2y) + (2y)^2 ) } =
2(x + 2y)(x^2 - 2xy + 4y^2)

Answer 17

x^3z^6−27y^9 @ranga

Answer 18

step by step please

Answer 19

@ranga

Answer 20

It is using the same principle again.
Make use of the identity: a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Answer 21

ok..

Answer 22

can you go step by step again?

Answer 23

I will do the first step. You should be able to do the remaining steps.
x^3z^6−27y^9 = (xz^2)^3 - (3y^3)^3
Make use of the identity: a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Here a = xz^2 and b = 3y^3. Just put a and b into the right hand side.

Answer 24

Im confused

Answer 25

@ranga

Answer 26

@ranga

Answer 27

x^3z^6−27y^9 = (xz^2)^3 - (3y^3)^3 = (xz^2 - 3y^3)(x^2z^4 - 3xy^3z^2 + 9y^6)