Question

how do i completely factor 25x^4 + 55x^2 - 60?

Answer 1

take x^2 as y and then substitute in the equation, you will get a quadratic equation in y solve it

Answer 2

at last after getting roots to y , then again use y = x^2 and get roots for x

Answer 3

Oh god i wish i was smart.

Answer 4

Someone help me. Im going to fail.

Answer 5

Well, I would consider factoring out a 5 from everything, since each term in the expression is divisible by 5.

Answer 6

iv gotten that far \[5x(5x^3-11x-12)\]

Answer 7

Hey @Studygirl01 im too like u , it takes little patience and time thats it

Answer 8

The 12 doesn't have an x so you can't factor a 5x from all three terms.

Answer 9

Oops, I meant the 60.

Answer 10

x^2=a---25*a^2+55*a-60=0 a(1,2)= \[\sqrt{a}=x\]

Answer 11

Once you factor out the 5, you need to find factors of -60 (the product of 5*(-12), which have a difference of -11. The factors of -60 which add to a -11 are -15 and +4, so replace your middle term with a -15x^2 and a +4x^2, and then you can factor by grouping.

Answer 12

ooops, I meant the factors of -60 which add to an 11, not a -11. The factors that will do that are a +15 and a -4.

Answer 13

So, then you can factor by grouping as folows:\[5(5x ^{4}+15x ^{2}-4x ^{2}-12)=5\left[ 5x ^{2}(x ^{2}+3)-4(x ^{2}+3) \right]\]

And then you can factor the common factor inside the brackets:

\[5\left[ (x ^{2}+3)(5x ^{2}-4) \right]\]

Answer 14

Thanks for the help, i'm just glad some people actually get this stuff, i'm still lost. I'l find someone to copy in the morning.