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Chapter 5: Problem 27

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{6 y^{3}-5 y^{2}+5}{3 y+2}$$

Short Answer

Expert verified
The quotient of the division is \(2y^2 - 3y\) and the remainder is \(6y + 5\). Verification confirmed that the solution is correct.

Step by step solution

01

Perform the division

Let's start the division. The first term in the dividend is \(6y^3\), and the first term in the divisor is \(3y\). So, the first term in the quotient is \(\frac{6y^3}{3y} = 2y^2\). Now, multiply the divisor by the first term of the quotient \(2y^2\). i.e., \((3y+2)(2y^2) = 6y^3 + 4y^2\), and then subtract this from the dividend which gives \(6y^3 - 4y^2 - (6y^3 + 4y^2) = -5y^2 - 4y^2 = -9y^2\).
02

Continue the division

The first term in the updated dividend is \(-9y^2\) and the first term in the divisor is \(3y\), so the next term in the quotient is \(\frac{-9y^2}{3y} = -3y\). Multiply the divisor by \(-3y\) and subtract this from the updated dividend, resulting: \(-9y^2 + 5 - (-9y^2 - 6y) = 6y + 5\). Now the dividend becomes a remainder as its degree is less than the degree of the divisor.
03

Write the final quotient and remainder

The final quotient has terms \(2y^2\) and \(-3y\) and the remainder is \(6y + 5\). So the division process can be written as \(6y^3 - 5y^2 + 5 = (3y + 2)(2y^2 - 3y) + (6y + 5)\).
04

Verify the solution

To verify the answer, multiply the divisor by the quotient and then add the remainder. Checking if this equals the original dividend, we get \((3y + 2)(2y^2 - 3y) + (6y + 5) = 6y^3 - 5y^2 + 5\). This is indeed equal to the original dividend \(6y^3 - 5y^2 + 5\), so the division is confirmed.

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Most popular questions from this chapter

Solve: \(4 x-7>9 x-2 .\) (Section 2.7, Example 7)

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