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Chapter 5: Problem 36

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{y^{4}-6 y^{2}+3}{y-1}$$

Short Answer

Expert verified
The quotient of \(y^{4}-6 y^{2}+3\) divided by \(y-1\) is \(y^{3}+ y^{2}+ y + 6\) and the remainder is 7.

Step by step solution

01

- Polynomial Long Division

Let's start performing the division \(\frac{y^{4}-6 y^{2}+3}{y-1}\). Begin by looking for the coefficient that will multiply the divisor \(y-1\) to give the first term of the dividend \(y^{4}\), this is \(y^{3}\). Multiply \(y-1\) by \(y^{3}\) and subtract the result from the dividend. The result is \(y^{3}-y^{2}\). Again multiply the divisor by the term which will give us \(y^{3}\), this time it is \(y^{2}\). Subtract the result \((y^{2}-y)\) from the previously found result \((y^{3}-y^{2})\). We get \(y^{2}-y+6\). Now, the term which will give us \(y^{2}\) is \(y\), subtract again. We finally get 7.
02

- Results of the Division

After these operations, the division will look as follows: the quotient is \(y^{3}+ y^{2}+ y + 6\) and the remainder is 7. This means \(\frac{y^{4}-6 y^{2}+3}{y-1} = y^{3}+ y^{2}+ y + 6 + \frac{7}{y-1}\)
03

- Verify the Results

Now that the division is finished, the goal is to check the results. This is done by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. Let's calculate it: \((y-1)*(y^{3}+ y^{2}+ y + 6)+7 = y^{4} - 6y^{2} + 3)\). This validates our results.

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