16 - a2
use the difference of two squares yule.
2(2x^2 + 2x - 5)
(2x + 5)(4x2 - 10x + 25)
if f(x) = x3 + 4x2 - 11x - 30, then yes, (x - 3) is indeed a factor. f(x) = x3 + 4x2 - 11x - 30 ∴ f(x) = (x - 3)(x2 + 7x + 10) ∴ f(x) = (x - 3)(x + 2)(x + 5)
4x2 - 9y Not factorable from what I see.
The GCF is 4x2.
4x2-y2 = (2x-y)(2x+y)
If that's + 4x2, the answer is 2x(x2 + 2x - 4)
4x2+20x+25 (2x+5)^2
4x2-36x+81 (2x-9)(2x-9)
4x2 - 23xy + 15y2 = (4x - 3y)(x - 5y)
You would factor it in this way: 2 * 2 * x * x.
4x
16 - a2
It is: 4x2-7x-15 when factored is (4x+5)(x-3)
Take out the common factor, which in this case is 4x2. Divide each of the terms by this common factor. 12x2 - 4x3 = 4x2(3+x)