# Solve: ` 8x^5+8x^4-64x^2-64x=0`

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### 1 Answer

`8x^5+8x^4-64x^2-64x=0`

Factor out the GCF of the forue terms which is 8x.

`8x(x^4+x^3-8x-8) = 0`

Then, to factor the expression inside the parenthesis, group the terms into two.

> `x^4+x^3-8x-8= (x^4+x^3)-(8x+8)`

Factor out the GCF in each group.

> `x^4+8x^3-8x-8=x^3(x+1)-8(x+1)`

> `x^4+8x^3-8x-8=(x+1)(x^3-8)`

So the equation becomes:

`8x(x+1)(x^3-8)=0`

Then, factor` x^3-8` , using the factor form of difference of two cubes which is `a^3-b^3=(a-b)(a^2+ab+b^2)` .

> `x^3-8=x^3-2^3=(x-2)(x^2+2x+2^2)=(x-2)(x^2+2x+4)`

So the factor of the given polynomial is:

`8x(x+1)(x-2)(x^2+2x+4)=0`

Then, set each factor equal to zero and solve for x.

> `8x=0`

`x=0`

> `x+1=0`

`x=-1`

> `x-2=0`

`x=2`

>`x^2+2x+4=0`

For the last factor, use the quadratic formula to solve for x.

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-2+-sqrt(2^2-4*1*4))/(2*1)`

`x=(-2+-sqrt(-12))/2=(-2+-2isqrt3)/2=-1+-isqrt3`

`x=-1+isqrt3 ` and `x=-1-isqrt3`

Hence, the solution to the equation is {`-1` , `0` , `2` , `-1+isqrt3` , `-1-isqrt3` }.