# How do you find the quotient of (14x^2+7x)div7x?

Jun 17, 2017

$2 x + 1$

#### Explanation:

You can compare this to

$3 x \left(5 x + 4\right) = 15 {x}^{2} + 12 x$

where the $\text{ "3x" }$ is multiplied by both terms inside the bracket using the distributive law,

$\left(14 {x}^{2} + 7 x\right) \div 7 x \text{ }$ can be written as $\text{ } \frac{1}{7 x} \left(14 {x}^{2} + 7 x\right)$

Or as $\text{ } \frac{14 {x}^{2} + 7 x}{7 x}$

Or as separate terms: $\text{ } \frac{14 {x}^{2}}{7 x} + \frac{7 x}{7 x}$

Whichever you choose, each of the terms has to be divided by $7 x$

$= 2 x + 1$

Jun 17, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{14 {x}^{2} + 7 x}{7 x} \implies \frac{14 {x}^{2}}{7 x} + \frac{7 x}{7 x} \implies \frac{14 {x}^{2}}{7 x} + 1$

Next, factor $7 x$ out of the numerator and denominator in the fraction on the left giving:

$\frac{7 x \times 2 x}{7 x \times 1} + 1 \implies \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7 x}}} \times 2 x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7 x}}} \times 1} + 1 \implies \frac{2 x}{1} + 1 \implies$

$2 x + 1$

However, from the original expression we cannot divide by $0$ therefore we need the exclusion:

Where: $7 x \ne 0$ or $x \ne 0$