Question

divide x(5x^2-80) by 5x(x+4)​

Answer 1

$\dfrac{x(5x^2-80)}{5x(x+4)}$ = x - 4

Step-by-step explanation:

We have,

$\dfrac{x(5x^2-80)}{5x(x+4)}$

To find, $\dfrac{x(5x^2-80)}{5x(x+4)}$ = ?

∴ $\dfrac{x(5x^2-80)}{5x(x+4)}$

= $\dfrac{(5x^2-80)}{5(x+4)}$

Taking 5 as common, we get

$=\dfrac{5(x^2-16)}{5(x+4)}$

$=\dfrac{(x^2-4^2)}{(x+4)}$

Using the algebraic identity,

$a^{2} -b^{2}$ =(a + b)(a - b)

$=\dfrac{(x+4)(x-4)}{(x+4)}$

= (x - 4)

∴ $\dfrac{x(5x^2-80)}{5x(x+4)}$ = x - 4