I'm having trouble figuring out how to factor this.

81y^4-72y^2+16

I know 16 is 2^4, so I rewrite as

81y^4-72y^2+2^4

I don't know where to go from here.

Thanks in advance.

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- Jul 30th 2011, 03:04 PMStudentMCCSHelp factoring something. What's the process/rule?
I'm having trouble figuring out how to factor this.

81y^4-72y^2+16

I know 16 is 2^4, so I rewrite as

81y^4-72y^2+2^4

I don't know where to go from here.

Thanks in advance. - Jul 30th 2011, 03:12 PMrdtedmRe: Help factoring something. What's the process/rule?
Use a substitution. (Turn the problem into a factorable quadratic)

let a = y^2

Then, 81a^2-72a+16 is the new equation using a (since a^2 = y^4)

Now to factor, you need two numbers whose product is (81x16=1296) and whose sum is -72 (-36 and -36)

Factor by grouping:

81a^2 - 36a - 36a + 16

9a(9a - 4) - 4(9a - 4)

(9a - 4)(9a - 4)

Resubstiture a = y^2

(9y^2-4)(9y^2-4) - Jul 30th 2011, 04:00 PMStudentMCCSRe: Help factoring something. What's the process/rule?
Thanks, I didn't know I could do that.

But the answer in the book is, (3y+2)^2(3y-2)^2. - Jul 30th 2011, 04:16 PMrdtedmRe: Help factoring something. What's the process/rule?
- Jul 30th 2011, 04:19 PMStudentMCCSRe: Help factoring something. What's the process/rule?
Thanks!

- Jul 30th 2011, 06:19 PMStudentMCCSRe: Help factoring something. What's the process/rule?
Hi again.

I've been wondering if there is an easy way to recognize the answer to this solution based on the relationships of its terms.

What I came up with is that.

Given AX^4-BX^2+C.

If (rootA)(rootB)(2)=B

Then AX^4-BX^2+C=, sorry I don't know how to properly write this out.

[(4th root A)X+(4th root C)]^2*[(4th root A)X-(4th root C)]^2

Is this really the case, I only checked for 2 equations? - Jul 30th 2011, 06:51 PMrdtedmRe: Help factoring something. What's the process/rule?
There's lots of "if's" to a generalization to this. If B < 0, for example, then simplest factored form is not written using 4th roots. Example:

If the original equation was

81y^4**+**72y^2+16 (B = -72)

Then it would factor to (9y^2 + 4)(9y^2 + 4)

Since both of the above factors have imaginary solutions, you would stop factoring here.

In short, you are correct, but occasionally taking this shortcut will yield a factored expression using imaginary factors.

Also, I would replace "If (rootA)(rootB)(2)=B" with "If (B/2)^2 = AC" - Jul 30th 2011, 07:01 PMStudentMCCSRe: Help factoring something. What's the process/rule?
Wait sorry, I meant (rootA)(RootC)(2)=B

- Jul 31st 2011, 03:58 AMSironRe: Help factoring something. What's the process/rule?
Is this the exercice:

Given $\displaystyle ax^4+bx^2+c$ (1)

If $\displaystyle 2\sqrt{a}\cdot \sqrt{b}=b$ (2)

then $\displaystyle ax^4+bx^2+c=...$?