How do you find all rational zeroes of the function using synthetic division $f(x)=x^4+x^3+x^2-9x-10$ ?

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Answer 1 · 2018-01-14T16:34:53

$x_1=-1-2i$ , $x_2=-1+2i$ , $x_3=-1$ and $x_4=2$

After using Rational Roots Test, $x=-1$ is a root of the polynomial. Hence $x+1$ is multiplier of it. Consequently,

$x^4+x^3+x^2-9x-10$

= $x^3*(x+1)+(x+1)*(x-10)$

= $(x+1)*(x^3+x-10)$

= $(x+1)*(x^3-8+x-2)$

= $(x+1)[(x-2)*(x^2+2x+4)+x-2]$

= $(x+1)(x-2)(x^2+2x+5)$

Hence roots of $f(x)$ are $-1, 2, -1+2i$ and $-1-2i$

How do you find all rational zeroes of the function using synthetic division f(x)=x^4+x^3+x^2-9x-10f(x)=x^4+x^3+x^2-9x-10?