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

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Answer 1 · 2016-03-18T19:11:40

$56x^3+43x^2+5x = x(7x+1)(8x+5)$

First separate out the common factor $x$ :

$56x^3+43x^2+5x = x(56x^2+43x+5)$

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

Find a pair of factors of $AC = 56*5 = 280$ with sum $B = 43$ .

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

Find the prime factorisation of $280$ :

$280 = 2*2*2*5*7$

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:

$1 + 5*7*2^3 = 1 + 280 = 281$

$5 + 7*2^3 = 5 + 56 = 61$

$7 + 5*2^3 = 7 + 40 = 47$

$5*7 + 2^3 = 35 + 8 = 43$

The last pair $35, 8$ works.

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

$56x^2+43x+5$

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

$=(56x^2+35x)+(8x+5)$

$=7x(8x+5)+1(8x+5)$

$=(7x+1)(8x+5)$

Putting it all together:

$56x^3+43x^2+5x = x(7x+1)(8x+5)$

How do you factor the expression 56x^3 +43x^2+5x56x^3 +43x^2+5x?