Question

Please help ASAP!!!!

Answer 1

Problem:

Remember :

If P(x) and D(x) are polynomials, then there exist unique polynomials Q(x) and R(x) such that:

`P(x)\text{ = D(x)Q(x)+ R(x)}`

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:

`P(x)\text{ = D(x)Q(x)+ R(x)}`

that:

`6y^2-11y+15\text{= (2y+7)(3y-16)+ }127`

Now, Dividing the previous expression by the divisor 2y+7, we obtain:

`\frac{6y^2-11y+15}{\text{2y+7}}\text{= 3y-16+ }\frac{127}{\text{2y+7}}`

then the correct answer is the first one:

`\text{3y-16+ }\frac{127}{\text{2y+7}}`