# How do you factor 3x^2-13x+4?

May 19, 2015

You can find its roots and then turn them into factors.

Using Bhaskara to find the roots:

$\frac{13 \pm \sqrt{169 - 4 \left(3\right) \left(4\right)}}{6}$
$\frac{13 \pm 11}{6}$
${x}_{1} = 4$, which is the same as the factor $x - 4 = 0$
${x}_{2} = \frac{1}{3}$, which is the same as the factor$3 x - 1 = 0$

$3 {x}^{2} - 13 x + 4 = \left(x - 4\right) \left(3 x - 1\right)$

May 19, 2015

There is another way. I use the new AC Method (Google, Yahoo Search) to factor trinomials

y = 3x^2 - 13x + 4 = (x - p)( - q)

Converted function: f'(x) = x^2 - 13x + 12.= (x - p')(x -q')

Compose factor pair of 12: (1, 12). This sum is 12 + 1 = 13 = -b.

Then p' = -1 and q' = -12.

Consequently, p = (p')/a = -1/3 and q' = (q')/a = -12/3 = -4

Factored form of y: y = (x - 1/3)(x - 4) = (3x - 1)(x - 4).