# How do you factor by grouping 30x^2 + 76x + 48 ?

Jun 7, 2015

$30 {x}^{2} + 76 x + 48$

$= 2 \left(15 {x}^{2} + 38 x + 24\right)$

$= 2 \left(15 {x}^{2} + 20 x + 18 x + 24\right)$

$= 2 \left(\left(15 {x}^{2} + 20 x\right) + \left(18 x + 24\right)\right)$

$= 2 \left(5 x \left(3 x + 4\right) + 6 \left(3 x + 4\right)\right)$

$= 2 \left(5 x + 6\right) \left(3 x + 4\right)$

How did I find a splitting of the middle $38 x$ term into $20 x + 18 x$ that would work?

The coefficients of $15 {x}^{2} + 38 x + 24$ are:

$A = 15$, $B = 38$, $C = 24$

We need to find a factorization of $A C = 15 \cdot 24 = 360$ into a pair of factors whose sum is $B = 38$.

$15 + 24 = 39$ is quite close to the $38$ we want, so the pair of numbers we're looking for will be similarly close to one another. In fact, notice that $361 = {19}^{2}$ and $19 + 19 = 38$ so
$\left(19 - 1\right) \left(19 + 1\right) = {19}^{2} - {1}^{2} = 361 - 1 = 360$ as required.

Jun 7, 2015

$f \left(x\right) = 2 \left(15 {x}^{2} + 38 x + 24\right) .$
If you like to avoid guessing, or avoid the lengthy factoring by grouping, use the systematic New AC Method.
$f \left(x\right) = 2 a \left(x - p\right) \left(x - q\right)$
Convert f(x) to f'(x) = x^2 + 38x + 360. = (x - p')(x - q'). Compose factor pairs of a.c = 360. a and c have same sign. Proceed:...(15, 24)(18, 20). OK. This sum is 18 + 20 = 38 = b. Then, p' = 18 and q' = 20
Then $p = \frac{p '}{a} = \frac{18}{15} = \frac{6}{5} , \mathmr{and} q = \frac{20}{15} = \frac{4}{3.}$

Factored form:$f \left(x\right) = 30 \left(x + \frac{4}{3}\right) \left(x + \frac{6}{5}\right) = 2 \left(3 x + 4\right) \left(5 x + 6\right)$