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Chapter 2: Problem 22

Use long division to divide. \(\left(x^{5}+7\right) \div\left(x^{3}-1\right)\)

Short Answer

Expert verified
\((x^{5}+7) \div (x^{3}-1) = x^{2}\), remainder \(7\)

Step by step solution

01

Set Up Division

Write the problem: \( (x^{5}+7) \div (x^{3}-1) \) in a long division setup, with \((x^{5}+7)\) inside the box and \((x^{3}-1)\) outside. Note that the polynomial inside the box is the dividend and the one outside the box is the divisor.
02

Divide the First Terms

Divide the first term of the dividend, \(x^{5}\), by the first term of the divisor, \(x^{3}\). The result (x^{2}) is the first term of the quotient.
03

Multiply and Subtract

Multiply the divisor by the term obtained in the quotient, subtract the result from the dividend, and bring down the next term.
04

Repeat the Process with the New Polynomial

Repeat the process of division and subtraction with the new polynomial until no terms are left to bring down.
05

Write the Final Result

The final result is the expression in the quotient box plus the remainder, divided by the divisor. If any term of the remainder has a degree lesser than that of the divisor then we stop the process. In this case, when we subtract, we get \(7\) which has a lesser degree than our divisor \(x^{3}-1\), so our result is \(x^{2}\) with a remainder of \(7\). Thus, \((x^{5}+7) \div (x^{3}-1) = x^{2}\) remainder \(7\).

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Most popular questions from this chapter

Between two consecutive zeros, a polynomial must be entirely ______________ or entirely __________________.

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