Question

. (5x2 - 16x - 21) divided by (x + 1)​

Answer 1

$\sf{Question}$

$\mathtt{(5x² - 16x - 21)÷(x + 1)}$

Step by step explaination:

$\implies{\tt{\dfrac{(5x² - 16x - 21)}{(x + 1)}}}$

$\implies{\tt{\dfrac{(5x²+5x−21x−21)}{(x + 1)}}}$

${\implies{\tt{\dfrac{5x(x+1)−21(x+1)}{(x + 1)}\: \: \: \: taking \ 5x \: and\: -21\: common}}}$

$\implies{\tt{ \dfrac{(5x−21)(x+1)}{(x + 1)}}}$

$\implies{\tt{\dfrac{(5x−21) \cancel{(x+1)}}{ \cancel{(x + 1)}}}}$

$\implies{\tt{ (5x−21)}}$

$\sf{ \pink{\therefore \ (5x−21)}\ wil\ get\ When\ we\ divide\ (5x² - 16x - 21) divided\ by\ (x + 1)}$

Answer 2

Answer:

Question

\mathtt{(5x² - 16x - 21)÷(x + 1)}(5x²−16x−21)÷(x+1)

Step by step explaination:

\implies{\tt{\dfrac{(5x² - 16x - 21)}{(x + 1)}}}⟹

(x+1)

(5x²−16x−21)

\implies{\tt{\dfrac{(5x²+5x−21x−21)}{(x + 1)}}}⟹

(x+1)

(5x²+5x−21x−21)

{\implies{\tt{\dfrac{5x(x+1)−21(x+1)}{(x + 1)}\: \: \: \: taking \ 5x \: and\: -21\: common}}}⟹

(x+1)

5x(x+1)−21(x+1)

taking 5xand−21common

\implies{\tt{ \dfrac{(5x−21)(x+1)}{(x + 1)}}}⟹

(x+1)

(5x−21)(x+1)

\implies{\tt{\dfrac{(5x−21) \cancel{(x+1)}}{ \cancel{(x + 1)}}}}⟹

(x+1)

(5x−21)

(x+1)

\implies{\tt{ (5x−21)}}⟹(5x−21)

\sf{ \pink{\therefore \ (5x−21)}\ wil\ get\ When\ we\ divide\ (5x² - 16x - 21) divided\ by\ (x + 1)}