How do you factor 7x^2 - 9x - 10?

Jun 22, 2016

$\left(7 x + 5\right) \left(x - 2\right)$

Explanation:

Let's start by finding the zeroes of the trynomial:

$x = \frac{9 \pm \sqrt{81 - 4 \left(7\right) \left(- 10\right)}}{2 \left(7\right)}$

$x = \frac{9 \pm \sqrt{81 + 280}}{4}$

$x = \frac{9 \pm \sqrt{361}}{14}$

$x = \frac{9 \pm 19}{14}$

$x = - \frac{10}{14} \mathmr{and} x = \frac{28}{14}$

that's

${x}_{1} = - \frac{5}{7} \mathmr{and} {x}_{2} = 2$

Since a trynomial can be factorized by the formula:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

so

$7 {x}^{2} - 9 x + 10 = 7 \left(x + \frac{5}{7}\right) \left(x - 2\right)$

or

$\left(7 x + 5\right) \left(x - 2\right)$

Jun 23, 2016

(7x + 5)(x - 2)

Explanation:

Use the new AC method to factor trinomials (Socratic Search)
$y = 7 {x}^{2} - 9 x - 10 =$ 7(x + p)(x + q)
Converted trinomial: $y ' = {x}^{2} - 9 x - 70 =$ (x + p')(x + q').
Find p' ans q' knowing they have opposite sign since ac < 0.
Factor pairs of (ac = -70) --> ...(-5, 14)(5, - 14). This sum is (-9 = b). Then, p' = 5 and q' = -14.
Back to original y --> $p = \frac{p '}{a} = \frac{5}{7}$, and $q = \frac{q '}{a} = - \frac{14}{7} = - 2$.
Factored form: $y = 7 \left(x + \frac{5}{7}\right) \left(x - 2\right) = \left(7 x + 5\right) \left(x - 2\right)$