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Chapter 2: Problem 5

In Exercises \(1-16,\) divide using long division. State the quotient, \(q(x),\) and the remainder, \(r(x)\). $$\left(6 x^{3}+7 x^{2}+12 x-5\right) \div(3 x-1)$$

Short Answer

Expert verified
The quotient \(q(x)\) is \(2x^{2} + 3x + 4\) and the remainder \(r(x)\) is 3.

Step by step solution

01

Setup

First, arrange the equation in the long division format with \(6 x^{3}+7 x^{2}+12 x-5\) as the dividend and \(3 x-1\) as the divisor.
02

Initial Division

Divide the leading term in the dividend \(6 x^{3}\) by the leading term in the divisor \(3x\). This gives us \(\frac{6 x^{3}}{3x} = 2x^{2}\), which is the first term of our quotient \(q(x)\).
03

Multiplication and Subtraction

Next, multiply the divisor \(3x-1\) by the first term of our quotient \(2x^{2}\). This gives \(6 x^{3}-2x^{2}\). Subtract this from our original dividend. After subtracting, we get \(7x^{2} + 2x^{2} + 12x - 5 = 9x^{2} + 12x - 5\) as the new dividend.
04

Repeating the Process

Repeat the process by dividing the leading coefficient of the new dividend by the leading coefficient of the divisor, adding to the quotient, multiplying, and subtracting in cycles until degrees of the dividend is less than the divisor. This gives us the quotient \(q(x) = 2x^{2} + 3x + 4\) and remainder \(r(x) = 3\).

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