# Find all the zeros of f(x) f(x) = 2x^4 - 7x^3 - 27x^2 + 63x + 81 List answers from smallest to largest. If there is a double root, list it twice.

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### 1 Answer

We have

`f(x)=2x^4-7x^3-27x^2+63x+81`

Apply remainder theorem

`f(-1)=2+7-27-63+81=0`

This means (x+1) is factor of f(x).

`f(x)=2x^4+2x^3-9x^3-9x^2-18x^2-18x+81x+81`

`=(x+1)(2x^3-9x^2-18x+81)`

`f(3)=(3+1)(54-81-54+81)=0`

This means (x-3) is facor of f(x)

`f(x)=(x+1)(2x^3-6x^2-3x^2+9x-27x+81)`

`f(x)=(x+1)(x-3)(2x^2-3x-27)`

factor

`2x^2-3x-27=2x^2-9x+6x-27`

`=x(2x-9)+3(2x-9)`

`=(x+3)(2x-9)`

`Thus`

`f(x)=(x+1)(x-3)(x+3)(2x-9)`

So zeros of f(x) are

-3,-1,3,9/2