How do you find all zeros with multiplicities of f(x)=x^3-7x^2+x-7?

Sep 17, 2017

The zeros of $f \left(x\right)$ are $7$, $i$, $- i$, all with multiplicity $1$.

Explanation:

Given:

$f \left(x\right) = {x}^{3} - 7 {x}^{2} + x - 7$

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

${x}^{3} - 7 {x}^{2} + x - 7 = \left({x}^{3} - 7 {x}^{2}\right) + \left(x - 7\right)$

$\textcolor{w h i t e}{{x}^{3} - 7 {x}^{2} + x - 7} = {x}^{2} \left(x - 7\right) + 1 \left(x - 7\right)$

$\textcolor{w h i t e}{{x}^{3} - 7 {x}^{2} + x - 7} = \left({x}^{2} + 1\right) \left(x - 7\right)$

Note that ${x}^{2} + 1$ has no linear factors with real coefficients since ${x}^{2} + 1 > 0$ for all real values of $x$. We can factor it as a difference of squares using complex coefficients:

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

So the zeros of $f \left(x\right)$ are $7$, $i$, $- i$, all with multiplicity $1$.