# How do you simplify (4w^2-49z^2)/(14z-4w)?

Jun 11, 2018

$= - \frac{2 w + 7 z}{2}$

#### Explanation:

Factorise both expressions

(4w^2-49z^2)/(14z-4w)" "(larr"difference of squares")/(" "larr "common factor")

$= \frac{\left(2 w + 7 z\right) \left(2 w - 7 z\right)}{2 \left(7 z - 2 w\right)}$

take out a common factor of $- 1$ from the denominator

$= \frac{\left(2 w - 7 z\right) \left(2 w + 7 z\right)}{- 2 \left(- 7 z + 2 w\right)} \text{ } \rightarrow = \frac{\left(2 w - 7 z\right) \left(2 w + 7 z\right)}{- 2 \left(2 w - 7 z\right)}$

$\frac{\cancel{\left(2 w - 7 z\right)} \left(2 w + 7 z\right)}{- 2 \cancel{\left(2 w - 7 z\right)}}$

$= - \frac{2 w + 7 z}{2}$

Jun 12, 2018

color(blue)(=> -(2w + 7z) / 2

#### Explanation:

$\frac{4 {w}^{2} - 49 {z}^{2}}{14 z - 4 w}$

Numerator is in the form $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

$\implies \frac{\left(2 w + 7 z\right) \left(2 w - 7 z\right)}{2 \cdot \left(7 z - 2 w\right)}$

$\implies - \frac{\left(2 w + 7 z\right) \cdot \cancel{7 z - 2 w}}{2 \cdot \cancel{7 z - 2 w}}$

$\implies - \frac{2 w + 7 z}{2}$