# How do you divide (x^4 + 3x^3 + 28x + 15) /( x + 5)?

Mar 17, 2016

$\frac{{x}^{4} + 3 {x}^{3} + 28 x + 15}{x + 5} = {x}^{3} - 2 {x}^{2} + 10 x - 22 + \frac{125}{x + 5}$

#### Explanation:

You can separate out multiples of $\left(x + 5\right)$ from the numerator progressively as follows:

$\frac{{x}^{4} + 3 {x}^{3} + 28 x + 15}{x + 5}$

$= \frac{{x}^{4} + 5 {x}^{3} - 2 {x}^{3} + 28 x + 15}{x + 5}$

$= \frac{{x}^{3} \left(x + 5\right) - 2 {x}^{3} + 28 x + 15}{x + 5}$

$= {x}^{3} + \frac{- 2 {x}^{3} - 10 {x}^{2} + 10 {x}^{2} + 28 x + 15}{x + 5}$

$= {x}^{3} - 2 {x}^{2} + \frac{10 {x}^{2} + 28 x + 15}{x + 5}$

$= {x}^{3} - 2 {x}^{2} + \frac{10 {x}^{2} + 50 x - 22 x + 15}{x + 5}$

$= {x}^{3} - 2 {x}^{2} + 10 x + \frac{- 22 x + 15}{x + 5}$

$= {x}^{3} - 2 {x}^{2} + 10 x + \frac{- 22 x - 110 + 125}{x + 5}$

$= {x}^{3} - 2 {x}^{2} + 10 x - 22 + \frac{125}{x + 5}$

This is equivalent to long division of polynomials.

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To check that the remainder is correct, substitute $x = - 5$ in the original numerator expression:

${x}^{4} + 3 {x}^{3} + 28 x + 15$

$= {5}^{4} - 3 \cdot {5}^{3} - 28 \cdot 5 + 15 = 625 - 375 - 140 + 15 = 125$

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If you prefer (as I do), you can long divide the coefficients - not forgetting to include a zero for the 'missing' ${x}^{2}$ term in the dividend...