Problem:
Remember :
If P(x) and D(x) are polynomials, then there exist unique polynomials Q(x) and R(x) such that:
[tex]P(x)\text{ = D(x)Q(x)+ R(x)}[/tex]
where
P(x) is the dividend
D(x) is the divisor
Q(x) is the quotient
R(x) is the remainder
Now, to solve the problem we will use the long division :
from the above, we have that
P(x) = the dividend = 6y^2 -11y+ 15
D(x) = the divisor = 2y + 7
Q(x) = the quotient = 3y-16
R(x) = the remainder = 127
then we can conclude from:
[tex]P(x)\text{ = D(x)Q(x)+ R(x)}[/tex]
that:
[tex]6y^2-11y+15\text{= (2y+7)(3y-16)+ }127[/tex]
Now, Dividing the previous expression by the divisor 2y+7, we obtain:
[tex]\frac{6y^2-11y+15}{\text{2y+7}}\text{= 3y-16+ }\frac{127}{\text{2y+7}}[/tex]
then the correct answer is the first one:
[tex]\text{3y-16+ }\frac{127}{\text{2y+7}}[/tex]
