Factor 16x^4 - 81Y^4 completely?

Question

I don't need the answer because if I wanted just the answer then I could get it out of the back of the textbook. I need to know the steps to obtain the answer. So, in order to make sure I am given the correct method, I won't put the answer on here. I'll choose the best answer according to who ever gives me the right answer with the method OR gives me the method that gets me the correct answer.

? · Accepted Answer

16x^4 - 81Y^4

 [ always remember : a^2-b^2=(a+b)(a-b)]


= (4x^2)^2-(9y^2)^2

= (4x^2+9y^2)(4x^2-9y^2)

= (4x^2+9y^2){(2x)^2-(3y)^2}

= (4x^2+9y^2)(2x+3y)(2x-3y) Ans.

? · Answer

This is a difference of squares, which means that a^2 - b^2 factors to (a + b)(a - b).
In this case, a = 4x^2 and b = 9y^2, so:

16x^4 - 81y^4 = (4x^2 + 9y^2)(4x^2 - 9y^2)

Now the second factor is still the difference of squares, with a = 2x and b = 3y, so factor further:

(4x^2 + 9y^2)(4x^2 - 9y^2) = (4x^2 + 9y^2)(2x + 3y)(2x - 3y)

It is now fully factored.

Answer:   (4x^2 + 9y^2)(2x + 3y)(2x - 3y)

? · Answer

Completely:

... 16x^4 - 81y^4 
= (4x^2)^2 - (9y^2)^2 
= (4x^2 - 9y^2) (4x^2 + 9y^2) 
= ( (2x)^2 - (3y)^2) ( (2x)^2 + (3y)^2) 
= ( 2x - 3y ) ( 2x + 3y ) ( 2x + 3y i ) ( 2x - 3y i )

Guillermo · Answer

(4*x^2 + 9*y^2)*(4*x^2 - 9*y^2) = (4*x^2 + 9*y^2)*(2*x + 3*y)*(2*x - 3*y)

Anonymous · Answer

difference of squares:

(4x^2-9y^2)(4x^2+9y^2)

then again:

(2x-3y)(2x+3y)(4x^2+9y^2)

iceman · Answer

(2x - 3y) (2x + 3y) (4x^2 + 9y^2)

Pranil · Answer

16x^4 - 81Y^4
= (4xÂ² + 9yÂ²)(4xÂ² – 9yÂ²
= (4xÂ² + 9yÂ²)(2x + 3y)(2x – 3y)
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Factor 16x^4 - 81Y^4 completely?

7 Answers