# How do you factor 24r^2 - 10r - 21?

May 23, 2015

$24 {r}^{2} - 10 r - 21$ is of the form $a {r}^{2} + b r + c$ with $a = 24$, $b = - 10$ and $c = - 21$.

The discriminant of this quadratic is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 10\right)}^{2} - \left(4 \times 24 \times - 21\right) = 100 + 2016 = 2116 = {46}^{2}$

Being a positive perfect square, the equation $24 {r}^{2} - 10 r - 21 = 0$ has two distinct rational roots, given by the formula:

$r = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{10 \pm 46}{48}$

That is $r = - \frac{36}{48} = - \frac{3}{4}$ and $r = \frac{56}{48} = \frac{7}{6}$.

Hence $24 {r}^{2} - 10 r - 21 = \left(4 r + 3\right) \left(6 r - 7\right)$

May 23, 2015

Factor f(x) = 24x^2 - 10x - 21 = (x - p)(x - q).
Use the new AC Method to factor trinomials. It is fast, systematic, no factoring, no solving binomials.
Converted trinomial $y ' = {x}^{2} - 10 x - 504$ = (x - p')(x - q').
Find p' and q' by composing factor pairs of a.c = -504. Proceed:... (-14, 16)(-18, 28). This last sum is 10 = -b. Then p' = 18 and q' = -28.

Then, $p = \frac{p '}{a} = \frac{18}{24} = \frac{3}{4}$, and $q = \frac{q '}{a} = - \frac{28}{24} = - \frac{7}{6}$

Factored form: $y = \left(x + \frac{3}{4}\right) \left(x - \frac{7}{6}\right) = \left(4 x + 3\right) \left(6 x - 7\right)$

Check by developing $y = 24 {x}^{2} - 28 x + 18 x - 21$ . OK