# How do you factor the expression 56x^3 +43x^2+5x?

Mar 18, 2016

$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(7 x + 1\right) \left(8 x + 5\right)$

#### Explanation:

First separate out the common factor $x$:

$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(56 {x}^{2} + 43 x + 5\right)$

To factor the remaining quadratic expression, use an AC method.

Find a pair of factors of $A C = 56 \cdot 5 = 280$ with sum $B = 43$.

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To help find the appropriate pair you can proceed as follow:

Find the prime factorisation of $280$:

$280 = 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7$

Next note that $43$ is odd, so it is the sum of an odd and an even number.

As a result, the prime factors must be split between the pair in such a way that all factors of $2$ are on one side or the other.

This leaves the following possibilities to check the sum:

$1 + 5 \cdot 7 \cdot {2}^{3} = 1 + 280 = 281$

$5 + 7 \cdot {2}^{3} = 5 + 56 = 61$

$7 + 5 \cdot {2}^{3} = 7 + 40 = 47$

$5 \cdot 7 + {2}^{3} = 35 + 8 = 43$

The last pair $35 , 8$ works.

Use this pair to split the middle term and factor by grouping:

$56 {x}^{2} + 43 x + 5$

$= 56 {x}^{2} + 35 x + 8 x + 5$

$= \left(56 {x}^{2} + 35 x\right) + \left(8 x + 5\right)$

$= 7 x \left(8 x + 5\right) + 1 \left(8 x + 5\right)$

$= \left(7 x + 1\right) \left(8 x + 5\right)$

Putting it all together:

$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(7 x + 1\right) \left(8 x + 5\right)$