Q. Divide. If there is a remainder, include it as a simplified fraction.(6y2+48y+42)÷(y+7)
Given Polynomial and Binomial: We are given the polynomial (6y2+48y+42) and we need to divide it by the binomial (y+7). We will use polynomial long division to solve this problem.
Dividing Leading Terms: First, we divide the leading term of the polynomial, 6y2, by the leading term of the binomial, y. This gives us 6y, because 6y2÷y=6y.
Multiplying and Subtracting: Next, we multiply the divisor (y+7) by the quotient we just found (6y) and subtract the result from the original polynomial. This gives us:(6y2+48y+42)−(6y×(y+7))=(6y2+48y+42)−(6y2+42y).
Dividing New Leading Term: Now, we divide the new leading term, 6y, by the leading term of the binomial, y. This gives us 6, because 6y÷y=6.
Multiplying and Subtracting: We multiply the divisor (y+7) by the new quotient (6) and subtract the result from the remaining polynomial. This gives us:(6y+42)−(6×(y+7))=(6y+42)−(6y+42).
Subtracting Result: Performing the subtraction, we get:(6y+42)−(6y+42)=(6y−6y)+(42−42)=0.
Exact Division: Since the remainder is 0, we do not have a remainder to express as a fraction. The division is exact.
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