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Please help ASAP!!!!

User Tekstrand
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1 Answer

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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}}

Please help ASAP!!!!-example-1
Please help ASAP!!!!-example-2
User Bpereira
by
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