Question

find the quotient and remainder using long division.
\frac{x^{3}-8 x^{2}+46}{x - 3}
the quotient is
the remainder is
points possible: 1
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Explanation:

Step1: Divide the leading term of the dividend by the leading term of the divisor

Divide \(x^{3}\) by \(x\). \(\frac{x^{3}}{x}=x^{2}\). Multiply \(x - 3\) by \(x^{2}\): \(x^{2}(x - 3)=x^{3}-3x^{2}\). Subtract this from \(x^{3}-8x^{2}+46\): \((x^{3}-8x^{2}+46)-(x^{3}-3x^{2})=- 5x^{2}+46\).

Step2: Repeat the process for the new dividend

Divide \(-5x^{2}\) by \(x\). \(\frac{-5x^{2}}{x}=-5x\). Multiply \(x - 3\) by \(-5x\): \(-5x(x - 3)=-5x^{2}+15x\). Subtract this from \(-5x^{2}+46\): \((-5x^{2}+46)-(-5x^{2}+15x)=-15x + 46\).

Step3: Repeat the process one more time

Divide \(-15x\) by \(x\). \(\frac{-15x}{x}=-15\). Multiply \(x - 3\) by \(-15\): \(-15(x - 3)=-15x+45\). Subtract this from \(-15x + 46\): \((-15x + 46)-(-15x+45)=1\).

Answer:

The quotient is \(x^{2}-5x - 15\).
The remainder is \(1\).