# How do you factor completely: 35x^2-54x+16?

Aug 8, 2015

Use quadratic formula to find roots of $35 {x}^{2} - 54 x + 16 = 0$ and deduce:

$35 {x}^{2} - 54 x + 16 = \left(5 x - 2\right) \left(7 x - 8\right)$

#### Explanation:

Let $f \left(x\right) = 35 {x}^{2} - 54 x + 16$

This is of the form $a {x}^{2} + b x + c$, with $a = 35$, $b = - 54$ and $c = 16$.

$f \left(x\right) = 0$ has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{54 \pm \sqrt{{\left(- 54\right)}^{2} - \left(4 \times 35 \times 16\right)}}{2 \times 35}$

$= \frac{54 \pm \sqrt{2916 - 2240}}{70}$

$= \frac{54 \pm \sqrt{676}}{70}$

$= \frac{54 \pm 26}{70}$

That is $x = \frac{28}{70} = \frac{2}{5}$ or $x = \frac{80}{70} = \frac{8}{7}$

So $\left(5 x - 2\right) = 0$ or $\left(7 x - 8\right) = 0$

Hence:

$35 {x}^{2} - 54 x + 16 = \left(5 x - 2\right) \left(7 x - 8\right)$

Aug 8, 2015

Factor $y = 35 {x}^{2} - 54 x + 16$

Ans: $\left(5 x - 2\right) \left(7 x - 8\right)$

#### Explanation:

I use the new AC Method to factor trinomials (Google, Yahoo Search)

$y = 35 {x}^{2} - 54 x + 16 =$35(x + p)(x + q)

Converted trinomial: $y ' = {x}^{2} - 54 x + 560.$
Find 2 numbers p' and q' knowing sum (-54 = b) and product (ac = 560)'. Here p' and q' have same sign.
Factor pairs of ac = 560 --> (10, 56)(14, 40). This sum is 54 = -b. Change the sum to the opposite.
Then p' = -14, and q' = -40, Therefore,
$p = \frac{p '}{a} = - \frac{14}{35} = - \frac{2}{5}$ and $q = \frac{q '}{a} = - \frac{40}{35} = - \frac{8}{7}$.

Factored form:

$y = 35 \left(x - \frac{2}{5}\right) \left(x - \frac{8}{7}\right) = \left(5 x - 2\right) \left(7 x - 8\right)$