Question

Divide. (20x^2-12x+8)/(2x+8)

Answer 1

your answer is might be wrong 460

Answer 2

Answer:

$10x-46+\frac{188}{x+4}$

Step-by-step explanation:

The given problem is :

$\frac{20x^{2}-12x+8}{2x+8}$

Taking out and cancelling 2 common from both numerator and denominator.

We get $\frac{10x^{2}-6x+4}{x+4}$

Divide the leading coefficient of numerator and divisor;

$\frac{10x^{2} }{x} =10x$

Quotient = 10x

Now, multiply x+4 by 10x = $10x^{2} +40x$

Subtracting $10x^{2} +40x$ from numerator above in question.

Remainder = $-46x+4$

Therefore, $\frac{10x^{2}-6x+4}{x+4}$ becomes:

$10x+\frac{-46x+4}{x+4}$

Again dividing the leading coefficient -46x by x; we get -46

Quotient = -46

Multiplying x+4 by -46, we get $-46x-184$

Subtracting $-46x-184$ from $-46x+4$ we get 188

So, remainder = 188

Therefore, the equation in final form becomes:

$10x-46+\frac{188}{x+4}$