# How do you factor the trinomial 8z^2+20z-48?

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#### Explanation:

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Apr 22, 2018

$4 \left(z + 4\right) \left(2 z - 3\right)$

#### Explanation:

you can divide all the terms by a common factor $4$ to get

$2 {z}^{2} + 5 z - 12$

this expression can be factorised by grouping:

the first and last coefficients are $2$ and $- 12$.
the product of this is $- 24$.

then, you can find $2$ numbers that add to give the second coefficient, and multiply to give the product of the first and last coefficient.

$8 + - 3 = 8 - 3 = 5$
$8 \cdot - 3 = - 24$

the two numbers that add to give the second coefficient are $8$ and $- 3$.
therefore the second term can be expressed as the sum of two terms, $8 z$ and $- 3 z$.

this gives the expression $2 {z}^{2} - 3 z + 8 z - 12$.

then you can factorise each adjacent pair:

$2 {z}^{2} - 3 z = z \left(2 z - 3\right)$
$8 z - 12 = 4 \left(2 z - 3\right)$

therefore $2 {z}^{2} - 3 z + 8 z - 12$ can be written as $z \left(2 z - 3\right) + 4 \left(2 z - 3\right)$.
this sum is the same as $\left(z + 4\right) \left(2 z - 3\right)$.

then you have the expression $2 {z}^{2} + 5 z - 12$ grouped as $\left(z + 4\right) \left(2 z - 3\right)$.

the expression given in the question is $8 {z}^{2} + 20 z - 48$.
that is $4$ times $2 {z}^{2} + 5 z - 12$, or the grouped expression $\left(z + 4\right) \left(2 z - 3\right)$.
therefore it can be written as $4 \left(z + 4\right) \left(2 z - 3\right)$.

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