# How do you factor 25x^2 - 10x + 4?

Jun 24, 2016

$25 {x}^{2} - 10 x + 4 = \left(5 x - 1 + i \sqrt{3}\right) \left(5 x - 1 - i \sqrt{3}\right)$

#### Explanation:

As the polynomial $25 {x}^{2} - 10 x + 4$ is quadratic, one can check discriminant to find if rational zeros and / or factors are possible or not. Discriminant of quadratic polynomial $\left(a {x}^{2} + b x + c\right)$ is $\left({b}^{2} - 4 a c\right)$. If it is negative, no real zeros / factors exist and if it is square of a number, rational zeros / factors are possible.

Here, $\left({b}^{2} - 4 a c\right) = \left({\left(- 10\right)}^{2} - 4 \times 25 \times 4\right) = \left(100 - 400\right) = - 300$ and as such no further real zeros / roots exist, but we can find complex binomials as factors.

As if $\alpha$ and $\beta$ are zeros of $a {x}^{2} + b x + c$, then $a {x}^{2} + b x + c = a \left(x - \alpha\right) \left(x - \beta\right)$ and hence let us find zeros of the trinomial using quadratic formula, which gives zeros as $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Hence, zeros of $25 {x}^{2} - 10 x + 4$ are

$\frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 25 \cdot 4}}{2 \cdot 25} = \frac{10 \pm \sqrt{- 300}}{50}$

i.e. $\frac{10 \pm 10 i \sqrt{3}}{50} = \frac{1 \pm i \sqrt{3}}{5}$

Hence $25 {x}^{2} - 10 x + 4 = 25 \left(x - \frac{1 - i \sqrt{3}}{5}\right) \left(x - \frac{1 + i \sqrt{3}}{5}\right)$

= $\left(5 x - 1 + i \sqrt{3}\right) \left(5 x - 1 - i \sqrt{3}\right)$