# How do you factor 7x^2 - 9x + 2?

Jun 6, 2015

You find its roots and turn them into factors by equaling them to zero.

Let's find the roots using Bhaskara:

$\frac{9 \pm \sqrt{81 - 4 \left(7\right) \left(2\right)}}{14}$

$\frac{9 \pm \sqrt{81 - 56}}{14}$

$\frac{9 \pm \sqrt{25}}{14}$

$\frac{9 \pm 5}{14}$

${x}_{1} = 1$, which, equaled to zero is $x - 1 = 0$
${x}_{2} = \frac{2}{7}$, which, equaled to zero is $7 x - 2 = 0$

Thus, we can factor $7 {x}^{2} - 9 x + 2$ this way:

$\left(x - 1\right) \left(7 x - 2\right)$

Jun 6, 2015

$f \left(x\right) = 7 {x}^{2} - 9 x + 2.$

Since (a + b + c) = 0, one factor is (x - 1) and the other is $\left(- \frac{c}{a} = - \frac{2}{7}\right)$

f(x) = (x - 1)(7x - 2)