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Chapter 5: Problem 20

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{9 y+18-11 y^{2}+12 y^{3}}{4 y+3}$$

Short Answer

Expert verified
The result of the polynomial division is a quotient of \(3y^{2} - y + 6\) and a remainder of \(2y + 12\). The check verifies that the result is correct: \((4y + 3)(3y^{2} - y + 6) + (2y + 12) = 12y^{3} - 11y^{2} + 9y + 18\)

Step by step solution

01

Arrange the polynomial and divisor for long division

Organize the polynomial \(12y^{3} - 11y^{2} + 9y + 18\) and the divisor \(4y + 3\) for long division. Order terms in descending powers of y.
02

Perform the Long division

Divide the first term of the dividend with the first term of the divisor to determine the first term of the quotient \(3y^{2}\). Multiply the entire divisor by \(3y^{2}\) and subtract the result from the dividend. Repeat this process for the new dividend \( -y^{2} + y + 18\)
03

Determine the remainder

After performing all possible divisions, the remaining polynomial \( 2y + 12 \) cannot be divided further by the divisor. This is the remainder.
04

Check the answer by reassembling the original polynomial

To check your answer, we can reassemble the original polynomial as \((divisor)*(quotient) + remainder\). When reassembled this way, we should get our original polynomial: \( (4y + 3)(3y^{2} - y + 6) + (2y + 12) = 12y^{3} - 11y^{2} + 9y + 18\)

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