# How do you factor 4x^2 - 6xy + 9y^2?

Aug 4, 2016

Factors of $4 {x}^{2} - 6 x y + 9 {y}^{2}$ are $\left(4 x - 3 y - 3 \sqrt{3} y i\right) \left(x - \frac{3}{4} y + \frac{3}{4} \sqrt{3} y i\right)$

#### Explanation:

$4 {x}^{2} - 6 x y + 9 {y}^{2}$ is a homogeneous quadratic function and factorizing it similar to those for quadratic function such as $a {x}^{2} + b x + c$, as taking ${y}^{2}$ out, makes it ${y}^{2} \left(4 {x}^{2} / {y}^{2} - \frac{6 x y}{y} ^ 2 + 9\right) = {y}^{2} \left(4 {\left(\frac{x}{y}\right)}^{2} - 6 \frac{x}{y} + 9\right) = {y}^{2} \left(4 {u}^{2} - 6 u + 9\right)$, where $u = \frac{x}{y}$.

As the discriminant ${b}^{2} - 4 a c = 36 - 144 = - 108$ is negative, we can only have complex factors.

Now zeros of $4 {x}^{2} - 6 x y + 9 {y}^{2}$ are $\frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \times 4 \times 9}}{2 \times 4}$ or $\frac{6 \pm \sqrt{- 108}}{8}$ or $\frac{3}{4} \pm \frac{3}{4} \sqrt{3} i$ and factors of $4 {u}^{2} - 6 u + 9$ could be

$4 \left(u - \frac{3}{4} - \frac{3}{4} \sqrt{3} i\right) \left(u - \frac{3}{4} + \frac{3}{4} \sqrt{3} i\right)$ or $\left(4 u - 3 - 3 \sqrt{3} i\right) \left(u - \frac{3}{4} + \frac{3}{4} \sqrt{3} i\right)$

Hence factors of $4 {x}^{2} - 6 x y + 9 {y}^{2}$ are $\left(4 x - 3 y - 3 \sqrt{3} y i\right) \left(x - \frac{3}{4} y + \frac{3}{4} \sqrt{3} y i\right)$