What are the zeroes of $f(x) = x^4 – x^3 + 7x^2 – 9x – 18$ ?

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Answer 1 · 2015-11-28T19:40:06

$x=-1$ , $x=3i$ , $x=-3i$ and $x=2$

First notice that by reversing the signs of the coefficients of the terms with odd degree, the sum is zero. So $x = -1$ is a zero:

$f(-1) = 1+1+7+9-18 = 0$

and $(x+1)$ is a factor:

$x^4-x^3+7x^2-9x-18$

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

then factor by grouping...

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

$=(x+1)(x^2(x-2)+9(x-2))$

$=(x+1)(x^2+9)(x-2)$

then take square root of $-9$ to find:

$=(x+1)(x-3i)(x+3i)(x-2)$

So the zeros are $x=-1$ , $x=3i$ , $x=-3i$ and $x=2$

What are the zeroes of f(x) = x^4 – x^3 + 7x^2 – 9x – 18f(x) = x^4 – x^3 + 7x^2 – 9x – 18?