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Chapter 2: Problem 14

Use long division to divide. \(\left(6 x^{3}-16 x^{2}+17 x-6\right) \div(3 x-2)\)

Short Answer

Expert verified
The quotient of \((6 x^{3}-16 x^{2}+17 x-6) \div(3 x-2)\) is \(2x^2 - 4x - 3\).

Step by step solution

01

Divide the leading term

Divide the leading term of the dividend, \(6x^3\), by the leading term of the divisor, \(3x\). This gives a quotient of \(2x^2\).
02

Subtract the product of the divisor and the quotient from the dividend

Multiply the divisor, \(3x - 2\), by the quotient from the previous step, \(2x^2\), which gives \(6x^3 - 4x^2\). Subtract this from the original dividend to get a new dividend \(6x^3 - 16x^2 + 17x - 6 - (6x^3 - 4x^2) = -12x^2 + 17x - 6\).
03

Repeat the process

Repeat the process starting from step 1 with the new dividend. Divide the leading term of the new dividend, \(-12x^2\), by the leading term of the divisor, \(3x\), which gives a quotient of \(-4x\). Then multiply this quotient by the divisor and subtract from the new dividend to get \( -12x^2 + 17x - 6 - (-4x(3x - 2)) = -9x + 6\).
04

Repeat the process one more time

Starting again from step 1, divide the leading term of the latest dividend \( -9x \) by the leading term of the divisor \( 3x \) to get a quotient of \(-3 \). Multiply it by the divisor and subtract it from the latest dividend to get a remainder of 0.
05

Combine the quotients

Combine the quotients from steps 1, 3, and 4 to get the final answer. The quotients were \(2x^2\), \(-4x\), and \(-3\), so the result of the division is \(2x^2 - 4x - 3\).

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