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Chapter 2: Problem 24

Use long division to divide. $$\left(5 x^{3}-16-20 x+x^{4}\right) \div\left(x^{2}-x-3\right)$$

Short Answer

Expert verified
The result of the division is \(x^{2}+6x-11\) with a remainder of \(7x-49\).

Step by step solution

01

Rearrange Terms

First, rearrange the terms in the dividend and the divisor in decreasing order by power. This gives \(x^{4}+5x^{3}-20x-16\) divided by \(x^{2}-x-3\).
02

Perform First Division

Next, divide the term with the highest power in the dividend (\(x^{4}\)) by the term with the highest power in the divisor (\(x^{2}\)). This gives \(x^{2}\). Write this above the division line.
03

Multiply and Subtract

Multiply the result \(x^{2}\) by every term in the divisor: \(x^{2}\cdot (x^{2}-x-3)=x^{4}-x^{3}-3x^{2}\). Subtract this from the original polynomial to find the remainder, which is \(6x^{3}-17x^{2}-16\).
04

Repeat Division

Repeat the process with the new polynomial. Divide the highest degree term of the new polynomial by the highest degree term of the divisor. This results in \(6x\). Multiply the divisor by \(6x\): \(6x\cdot (x^{2}-x-3)=6x^{3}-6x^{2}-18x\). Subtract this from \(6x^{3}-17x^{2}-16\) to get the new remainder \(-11x^{2}+18x-16\).
05

Final Division

Repeat the process one more time. Divide the highest degree term of the new polynomial by the highest degree term of the divisor. This results in \(-11\). Multiply the divisor by \(-11\):\(-11\cdot (x^{2}-x-3)=-11x^{2}+11x+33\), and subtract this from \(-11x^{2}+18x-16\) to get the remainder \(7x-49\). Since the degree of this remainder is less than the degree of the divisor, we can stop here.

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Most popular questions from this chapter

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