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.
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)
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?
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"