# How do you find the zeroes of  f(x)= x^3 - 17x^2 + 81x -65?

Mar 3, 2018

Zeros are $x = 1 , x = 8 - 1 i \mathmr{and} x = 8 + 1 i$

#### Explanation:

$f \left(x\right) = {x}^{3} - 17 {x}^{2} + 81 x - 65$ or

$f \left(x\right) = {x}^{3} - {x}^{2} - 16 {x}^{2} + 16 x + 65 x - 65$ or

$f \left(x\right) = {x}^{2} \left(x - 1\right) - 16 x \left(x - 1\right) + 65 \left(x - 1\right)$ or

$f \left(x\right) = \left(x - 1\right) \left({x}^{2} - 16 x + 65\right) \therefore x = 1$ is one zero

$f \left(x\right) = {x}^{2} - 16 x + 65$ or

$f \left(x\right) = {x}^{2} - 16 x + 64 + 1$ or

$f \left(x\right) = {\left(x - 8\right)}^{2} - \left({i}^{2}\right) \left[{i}^{2} = - 1\right]$ or

$f \left(x\right) = \left(x - 8 + i\right) \left(x - 8 - i\right) \therefore$. Other zeros are

$x = 8 - 1 i \mathmr{and} x = 8 + 1 i$

Zeros are $x = 1 , x = 8 - 1 i \mathmr{and} x = 8 + 1 i$ [Ans]