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

1 Answer
Mar 18, 2016

Answer:

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

Explanation:

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)#