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Chapter 5: Problem 30

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{8 x^{3}+27}{2 x+3}$$

Short Answer

Expert verified
The quotient is \(4x^2 - 6x + 9\) and the remainder is 0. This is verified by showing that \((4x^2 - 6x + 9)(2x + 3) + 0 = 8x^3 + 27\), the original dividend.

Step by step solution

01

Begin the Polynomial Long Division

Let's proceed with polynomial long division. We'll divide \(8 x^{3}+27\) by \(2x+3\). Since the leading term of the divisor divides \(8x^3\), use that to determine the first term of the quotient, which is \(4x^2\). Multiply \(4x^2\) with \(2x+3\) which results in \(8x^3 + 12x^2\). Subtract this from the original polynomial to obtain the new polynomial to be divided.
02

Continue the Polynomial Long Division

Subtracting \(8x^3 + 12x^2\) from the original polynomial, we obtain \(-12x^2 + 27\). Again, divide the first term of that polynomial by the first term of the divisor, which gives us \(-6x\). Multiply \(-6x\) with \(2x + 3\) to get \(-12x^2 -18x\). Subtract this from \(-12x^2 + 27\) to get \(18x + 27\).
03

Continue the Polynomial Long Division until the degree of the remainder is less than the degree of the divisor.

Divide \(18x\) by \(2x\), which gives us \(9\). We multiply \(9\) by \(2x + 3\) to obtain \(18x + 27\). Subtract this from \(18x + 27\), which gives us the remainder of 0.
04

Verify the division

To verify the division, we should regain the original dividend when multiplying the quotient \((4x^2 -6x + 9)\) with the divisor \((2x + 3)\) and adding the remainder. Doing this multiplication and addition, we obtain \(8x^3 + 27\), which is indeed our original dividend.

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Most popular questions from this chapter

In Exercises \(100-103,\) determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If a polynomial in \(x\) of degree 6 is divided by a monomial in \(x\) of degree \(2,\) the degree of the quotient is 4

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