Answer:
[tex]10x-46+\frac{188}{x+4}[/tex]
Step-by-step explanation:
The given problem is :
[tex]\frac{20x^{2}-12x+8}{2x+8}[/tex]
Taking out and cancelling 2 common from both numerator and denominator.
We get [tex]\frac{10x^{2}-6x+4}{x+4}[/tex]
Divide the leading coefficient of numerator and divisor;
[tex]\frac{10x^{2} }{x} =10x[/tex]
Quotient = 10x
Now, multiply x+4 by 10x = [tex]10x^{2} +40x[/tex]
Subtracting [tex]10x^{2} +40x[/tex] from numerator above in question.
Remainder = [tex]-46x+4[/tex]
Therefore, [tex]\frac{10x^{2}-6x+4}{x+4}[/tex] becomes:
[tex]10x+\frac{-46x+4}{x+4}[/tex]
Again dividing the leading coefficient -46x by x; we get -46
Quotient = -46
Multiplying x+4 by -46, we get [tex]-46x-184[/tex]
Subtracting [tex]-46x-184[/tex] from [tex]-46x+4[/tex] we get 188
So, remainder = 188
Therefore, the equation in final form becomes:
[tex]10x-46+\frac{188}{x+4}[/tex]
