How do i factor x^4 - 13x^2 + 36??

Answer 1

x^4 - 13x^2 + 36
= (x^2 - 4) (x^2 - 9)
= (x - 2) (x + 2) (x - 3) (x + 3)

By the rule:
(a + b) (a - b) = a^2 - b^2

Answer 2

well, first you have to realize what factors of 36 add up to 13? the factors of 36 are 1,2,3,4,6,9,18,36. So which of those add up to 13?-4 and 9. since the 13 in the problem is negative, and the 36 is positive, your problem has to look like this: (x-a)(x-b). since x^2 times x^2 is x^4, you know what your "x" is going to be. and your a will be 9 or 4 and the same for your b. if you're in doubt, check by using the FOIL method. Good Luck!

Answer 3

Treat it like this quadratic:

y^2 - 13y + 36

Because your question is the same as
(x^2)^2 - 13x^2 + 36

With that in mind, this factors to

(x^2 - 9)(x^2 - 4)

And note that we have two difference of squares, which can be further factored as

(x - 3)(x + 3)(x - 2)(x + 2)

Answer 4

Imagine that it is a quadratic:

x² - 13x + 36

The solution would be (x - 9)(x - 4)

Just replace x by x²

Solution is (x² - 9)(x² - 4)

Now factor these:

((x + 3)(x - 3)(x + 2)(x - 2)

Answer 5

-9 times -4 is 36
-9-4=-13
(x-9)(x-4)

Answer 6

x^4-13x^2+36
=x^4-9x^2-4x^2+36
=x^2(x^2-9)-4(x^2-9)
=(x^2-4)(x^2-9)
=(x^2-2^2)(x^2-3^2)
=(x-2)(x+2)(x-3)(x+3)-----using a^2-b^2=(a-b)(a+b) ans

Answer 7

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