# How do you find the zeroes of  f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25?

Nov 27, 2015

Factor by grouping twice, then use the difference of squares identity twice and hence find zeros:

$\pm \frac{\sqrt{10}}{2}$, $\pm \sqrt{5}$, $- \frac{1}{5}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Factor by grouping:

$f \left(x\right) = 10 {x}^{5} + 2 {x}^{4} - 75 {x}^{3} - 15 {x}^{2} + 125 x + 25$

$= \left(10 {x}^{5} + 2 {x}^{4}\right) - \left(75 {x}^{3} + 15 {x}^{2}\right) + \left(125 x + 25\right)$

$= 2 {x}^{4} \left(5 x + 1\right) - 15 {x}^{2} \left(5 x + 1\right) + 25 \left(5 x + 1\right)$

$= \left(2 {x}^{4} - 15 {x}^{2} + 25\right) \left(5 x + 1\right)$

$= \left(2 {x}^{4} - 10 {x}^{2} - 5 {x}^{2} + 25\right) \left(5 x + 1\right)$

$= \left(\left(2 {x}^{4} - 10 {x}^{2}\right) - \left(5 {x}^{2} - 25\right)\right) \left(5 x + 1\right)$

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

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

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

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

Hence $f \left(x\right)$ has zeros for the following values of $x$

$\pm \frac{\sqrt{5}}{\sqrt{2}} = \pm \frac{\sqrt{10}}{2}$

$\pm \sqrt{5}$

$- \frac{1}{5}$