# How do you factor 49a^2 - 112a + 16?

Jun 8, 2016

49a^2-112a+16=(7a-8-4sqrt3))(7a-8+4sqrt3))

#### Explanation:

In $a {x}^{2} + b x + c$, to factorize we split the middle term $b$ in two parts so that they add up to $b$ and their product is $a c$. But for this discriminant ${b}^{2} - 4 a c$ has to be square of a rational number.

Here in $49 {a}^{2} - 112 a + 16$, we have to split $- 112$ in two parts so that their product is $16 \times 49$. But discriminant is
${112}^{2} - 4 \times 49 \times 16 = 12544 - 3136 = 9408$, which is not the square of a rational number. Hence we cannot split it to factorize it.

How can then we factorize $49 {a}^{2} - 112 a + 16$? This can be done using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, which gives zeros of function. Here these are

$\frac{- \left(- 112\right) \pm \sqrt{{112}^{2} - 4 \times 49 \times 16}}{2 \cdot 49} = \frac{112 \pm \sqrt{9408}}{98}$

= $\frac{112 \pm 56 \sqrt{3}}{98} = \frac{8 \pm 4 \sqrt{3}}{7}$

Hence factors of $49 {a}^{2} - 112 a + 16$ are $49 \left(a - \frac{8 + 4 \sqrt{3}}{7}\right) \left(a - \frac{8 - 4 \sqrt{3}}{7}\right)$ or

(7a-8-4sqrt3))(7a-8+4sqrt3))