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Chapter 2: Problem 21

Use long division to divide. \(\left(x^{3}-9\right) \div\left(x^{2}+1\right)\)

Short Answer

Expert verified
The quotient is \(x\) with a remainder of \(x^{2} - 9\)

Step by step solution

01

Set Up the Long Division

First, set up the long division, similar as you would do for numerical long division. Write \(x^{3}-9\) inside the division bracket and \(x^{2}+1\) outside. Remember that if either polynomial is missing a degree of \(x\), you should use 0 as a placeholder for that degree of \(x\) in the polynomial. Thus, you should rewrite \(x^{3}-9\) as \(x^{3}+ 0x^{2} + 0x - 9\).
02

Start the Division

Divide the first term from the dividend (\(x^{3}\)) by the first term of the divisor (\(x^{2}\)). This will give you the first term of the quotient, \(x\). Multiply the divisor by this term (\(x\)), and subtract the result from the dividend. The result is \(x^{2} - 9\). Remember to change the signs when subtracting this polynomial.
03

Continue the Division

Repeat the process with this new dividend (\(x^{2} - 9\)). Divide the first term in this new dividend by the first term of the divisor to get the next term in the quotient (no value, since \(x^{2} / x^{2}\) is just 1 and when subtracted from first term in the quotient, which is \(x\), will remain \(x\)). Subtract the result of the multiplication from the new dividend. As you can't subtract 1 from \(x\), it will show that the divisor cannot anymore be subtracted from dividend and, hence, the division stops here.
04

Write down the Remainder

The polynomial that's left is the remainder. Here, it is \(x^{2} - 9\). The original division problem can thus be rewritten as \(x^{3} - 9 = (x^{2} + 1)q + r, where q is the quotient and r is the remainder.

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