Question

4x^3-25x and 4x^3-7x^2-x they're both factorization equations, how do you solve them?

Answer 1

4x^3-25x=x(4x^2-25)=0

Answer 2

4x^3-7x^2-x

Answer 3

do the same

Answer 4

\[4x^3 - 25x = {x}(4x^2 - 25) = x({(2x)^2-(5^2))}=x(2x+5)(2x-5)\]

Answer 5

so now we need to factorize

Actually whenever if any equation ends at x that means there is a root x=0

so you have to take x common and then put the other part =0

Answer 6

\[4x^3 - 7x^2 -x = {x}(4x^2 - 7x-1)\]