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

Jan 14, 2018

${x}_{1} = - 1 - 2 i$, ${x}_{2} = - 1 + 2 i$, ${x}_{3} = - 1$ and ${x}_{4} = 2$

Explanation:

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

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

=${x}^{3} \cdot \left(x + 1\right) + \left(x + 1\right) \cdot \left(x - 10\right)$

=$\left(x + 1\right) \cdot \left({x}^{3} + x - 10\right)$

=$\left(x + 1\right) \cdot \left({x}^{3} - 8 + x - 2\right)$

=$\left(x + 1\right) \left[\left(x - 2\right) \cdot \left({x}^{2} + 2 x + 4\right) + x - 2\right]$

=$\left(x + 1\right) \left(x - 2\right) \left({x}^{2} + 2 x + 5\right)$

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