Question

find he zero for each function y=x^4-2x^3-x^2-2x

Answer 1

y=x^4-2x^3-x^2-2x

factor by grouping

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

x^3(x - 2) - x( x - 2)

(x^3-x)(x-2)
x(x^2-1)(x-2)
x(x+1)(x-1)(x-2)
x=0,1,-1,2

what i thought lol

Answer 2

Did the sign of the -2x get changed in the solution. Looking at the third line at
"-(x^2 -2x)" changes the sign of 2x It is a negative in the original equation.

Answer 3

i didn't factor out "-x", if that's what you are going for. i chose to factor out a positive x to keep both factors of (x-2) on the left and (x-2) on the right equal.

Answer 4

When the factors x(x+1)(x-1)(x-2) are multiplied the results give you
x^4-2x^3-x^2+2x
Check it out.

Answer 5

even if you do pull out a negative x, and pull out a -x^2 likewise from the other side, the answer stays the same:

y=x^4-2x^3-x^2-2x

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

(x-x^3)(2-x)
-x(x^2-1)(2-x)
x(x^2-1)(x+2)
x(x+1)(x-1)(x+2)

x=0,-1,1,-2

Answer 6

hmm lemme see

Answer 7

i got a different answer that second time actually - (x+2), which is the needed term to make the -2x.

(x-2)(x^2 - 1) = neg*neg = pos 2x
(x+2)(x^2 - 1) = pos*neg = neg 2x

apparently i dont check my work close enough lol. thanks for looking at that lol
!

Answer 8

Is 1 a solution? Does it result in a 0? The -2 and 0 are solutions.

Answer 9

I'm checking the -1 now for a zero.

Answer 10

The -1 solution results in 4.

Answer 11

To be honest, I think the problem was meant to be a +2x for the last term. lol

Answer 12

@cherry33, do you have a solution sheet?

Answer 13

And have you copies the equation correctly?