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

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

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Answer 1 · 2016-03-18T19:11:40

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

Explanation:

First separate out the common factor $x$ $x$ :

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

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

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

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

Find the prime factorisation of $280$ $280$ :

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

Next note that $43$ $43$ 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 $2$ $2$ are on one side or the other.

This leaves the following possibilities to check the sum:

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

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

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

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

The last pair $35, 8$ $35, 8$ works.

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

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

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

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

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

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

Putting it all together:

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

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

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