# How do you divide (15x^2 + 22x + 8)/( 5x + 4)?

Dec 27, 2015

$3 x + 2$

There is a slight problem with this. Look at the footnote in the explanation.

#### Explanation:

Method 1: See if you can factorise the numerator such that it cancels out part of the denominator

Method 2: Long division.

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$\textcolor{b l u e}{\text{Trying method 1 first}}$

Factors of 15 -> {1,15} ; {3,5}
Factors of 8 -> {1,8} ; {2,4}

Attempt No 1: For this to work we need $\left(5 x + 4\right)$

=> (5x+4)(?_1 x+?_2) =15x^2+22x+8

Using our list of factors this would give us:

$\left(5 x + 4\right) \left(3 x + 2\right) = 15 {x}^{2} + 10 x + 12 x + 8$
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This works so we have:

$\frac{\left(5 x + 4\right) \left(3 x + 2\right)}{\left(5 x + 4\right)}$

$\frac{5 x + 4}{5 x + 4} \times \left(3 x + 2\right)$

But $\frac{5 x + 4}{5 x + 4} = 1$ giving

$3 x + 2$ as the answer.

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However, you have to be mindful of the fact that division by 0 is 'not allowed' in mathematics (Undefined). Thus there is a value of x where $\left(5 x + 4\right) = 0$ so division by $\left(5 x + 4\right)$ at that point gives you a bit of a problem!