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Divide. If there is a remainder, include it as a simplified fraction.\newline(6y2+48y+42)÷(y+7)(6y^2 + 48y + 42) \div (y + 7)

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Q. Divide. If there is a remainder, include it as a simplified fraction.\newline(6y2+48y+42)÷(y+7)(6y^2 + 48y + 42) \div (y + 7)
  1. Given Polynomial and Binomial: We are given the polynomial (6y2+48y+42)(6y^2 + 48y + 42) and we need to divide it by the binomial (y+7)(y + 7). We will use polynomial long division to solve this problem.
  2. Dividing Leading Terms: First, we divide the leading term of the polynomial, 6y26y^2, by the leading term of the binomial, yy. This gives us 6y6y, because 6y2÷y=6y6y^2 \div y = 6y.
  3. Multiplying and Subtracting: Next, we multiply the divisor (y+7)(y + 7) by the quotient we just found (6y)(6y) and subtract the result from the original polynomial. This gives us:\newline(6y2+48y+42)(6y×(y+7))=(6y2+48y+42)(6y2+42y)(6y^2 + 48y + 42) - (6y \times (y + 7)) = (6y^2 + 48y + 42) - (6y^2 + 42y).
  4. Subtracting Result: Performing the subtraction, we get: \newlineegin{equation}\newline(66y^22 + 4848y + 4242) - (66y^22 + 4242y) = (66y^22 - 66y^22) + (4848y - 4242y) + 4242 = 66y + 4242.\newlineegin{equation}
  5. Dividing New Leading Term: Now, we divide the new leading term, 6y6y, by the leading term of the binomial, yy. This gives us 66, because 6y÷y=66y \div y = 6.
  6. Multiplying and Subtracting: We multiply the divisor (y+7)(y + 7) by the new quotient (6)(6) and subtract the result from the remaining polynomial. This gives us:\newline(6y+42)(6×(y+7))=(6y+42)(6y+42)(6y + 42) - (6 \times (y + 7)) = (6y + 42) - (6y + 42).
  7. Subtracting Result: Performing the subtraction, we get:\newline(6y+42)(6y+42)=(6y6y)+(4242)=0(6y + 42) - (6y + 42) = (6y - 6y) + (42 - 42) = 0.
  8. Exact Division: Since the remainder is 00, we do not have a remainder to express as a fraction. The division is exact.

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