How do you find all the zeros of $f(x) = x^4 – x^3 + 7x^2 – 9x – 18$ ?

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Answer 1 · 2016-05-30T09:15:54

$p_4(x)=(x+1)(x-2)(x+3i)(x-3i)$

The constant coefficient divided by the maximum power coefficient in a polynomial, gives the product of all its roots.

So $18/1=x_1 x_2 x_3 x_4$ for a polynomial represented as

$p_4(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$

Supposing integer roots and testing for $pm 1,pm2,pm3,pm6$
we obtain:

$p_4(-1)=p_4(2)=0$ so we have at least a root for each $1$ and $-1$
then we can do

$p_4(x)=(x+1)(x-2)(x-x_3)(x-x_4)=x^4-x^3+7x^2-9x-18$

Now dividing $(x^4-x^3+7x^2-9x-18)/((x+1)(x-2)) = x^2+9$

so all the roots are

$p_4(x)=(x+1)(x-2)(x+3i)(x-3i)$

How do you find all the zeros of f(x) = x^4 – x^3 + 7x^2 – 9x – 18f(x) = x^4 – x^3 + 7x^2 – 9x – 18?