Question

divide. if there is a remainder, include it as a simplified fraction. (6g³ - 20g² + 6g) ÷ (g - 3)

Explanation:

Step1: Use polynomial long division. Divide the first term of the dividend by the first term of the divisor.

Divide \(6g^3\) by \(g\) to get \(6g^2\). Multiply the divisor \((g - 3)\) by \(6g^2\) to get \(6g^3 - 18g^2\). Subtract this from the dividend:

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Step2: Divide the first term of the new dividend by the first term of the divisor.

Divide \(-2g^2\) by \(g\) to get \(-2g\). Multiply the divisor \((g - 3)\) by \(-2g\) to get \(-2g^2 + 6g\). Subtract this from the new dividend:

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Wait, that can't be right. Wait, let's check again. Wait, maybe I made a mistake. Wait, let's redo the subtraction. Wait, the original dividend is \(6g^3 - 20g^2 + 6g\). After the first step, we had \(-2g^2 + 6g\). Now, divide \(-2g^2\) by \(g\) to get \(-2g\). Multiply \((g - 3)\) by \(-2g\) gives \(-2g^2 + 6g\). Subtract that from \(-2g^2 + 6g\) gives 0. But wait, that would mean the division is exact? But let's check by plugging \(g = 3\) into the original polynomial. \(6(3)^3 - 20(3)^2 + 6(3) = 627 - 209 + 18 = 162 - 180 + 18 = 0\). Oh, so the remainder is 0. Wait, but let's check the coefficients again. Wait, \(6g^3 - 20g^2 + 6g\) divided by \(g - 3\). Let's use synthetic division. The root is \(g = 3\). Coefficients: 6, -20, 6, 0 (since there's no constant term). Bring down the 6. Multiply by 3: 18. Add to -20: -2. Multiply by 3: -6. Add to 6: 0. Multiply by 3: 0. Add to 0: 0. So the quotient is \(6g^2 - 2g + 0\), so \(6g^2 - 2g\). Wait, that makes sense. So the division is exact.
Wait, let's verify by multiplying \((g - 3)(6g^2 - 2g)\). \(g*6g^2 = 6g^3\), \(g(-2g) = -2g^2\), \(-3*6g^2 = -18g^2\), \(-3(-2g) = 6g\). Combine like terms: \(6g^3 - 2g^2 - 18g^2 + 6g = 6g^3 - 20g^2 + 6g\), which matches the dividend. So the quotient is \(6g^2 - 2g\) with a remainder of 0.

Wait, but earlier when I did the long division, after the first two steps, the remainder was 0. So the result is \(6g^2 - 2g\).

Wait, let's check again. \( (g - 3)(6g^2 - 2g) = 6g^3 - 2g^2 - 18g^2 + 6g = 6g^3 - 20g^2 + 6g \), which is the original dividend. So yes, the division is exact.

Answer:

\(6g^2 - 2g\)