Problem 20 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The quotient when \(x^{5}+7\) is divided by \(x^{3}-1\) is \(x^{2}\), and the remainder is \(x^{2}+7\). Therefore, \((x^{5}+7)\) \div \((x^{3}-1)\) = \(x^{2}\)

Step 1: Setup the Long Division

Arrange \(x^{5}+7\) under the division bar and \(x^{3}-1\) to the left of the division bar. Remember to fill in any missing terms with a zero coefficient in the dividend polynomial i.e. \(x^{5}+0x^{4}+0x^{3}+0x^{2}+0x+7\).

Step 2: Divide the First Terms

Take the first term of the polynomial inside the bracket and divide it by the first term of the polynomial outside the bracket. In other words, divide \(x^{5}\) by \(x^{3}\). This gives \(x^{2}\). Write \(x^{2}\) above the division bar.

Step 3: Multiply and Subtract

Multiply \(x^{2}\) (found in Step 2) by the polynomial outside the bracket \(x^{3}-1\). This will give \(x^{5}-x^{2}\). Subtract this from the polynomial inside the bracket. This is done by subtracting each term individually i.e. \(x^{5} - x^{5}\) and \(0x^{4}-(-x^{2})\). This results in \(x^{2}\). This process is repeated until we can't get any terms with a higher degree than the polynomial outside the bracket.

Step 4: Repeat Division

Repeat the steps above until the degree of the remaining polynomial is less than the degree of the divisor, in this case \(x^{3}-1\). The equation becomes \(x^{2}\), which is less than the degree of the divisor \(x^{3}-1\). So we stop here.