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Chapter 5: Problem 3

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{2 x^{2}+x-10}{x-2}$$

Short Answer

Expert verified
The quotient is \(2x - 3\) with a remainder of \(-16\), therefore the result of the division is \(2x - 3 - \frac{16}{x-2}\).

Step by step solution

01

Perform the Polynomial Division

Follow the method of polynomial division. Divide the first term of the dividend \((2x^{2})\) by the first term of the divisor \((x)\) to get \(2x\). Now subtract the product of the divisor and \(2x\) from the dividend and bring down the next term of the dividend.\nThis gives \(2x^{2} + x - 2(2x^{2} - 4x) = -3x - 10\).
02

Continue the Division

We continue dividing. We now divide \(-3x\) by \(x\), the first term of the divisor, giving us \(-3\). Subtract the product of the divisor and \(-3\) from \(-3x - 10\), which results in \(-3x - 10 - (-3x + 6) = -16\).
03

Write the Final Answer

As degree of remaining polynomial is less than the degree of the divisor, it is the remainder. Therefore, the quotient when \(2x^{2} + x - 10\) is divided by \(x - 2\) is \(2x - 3\) and the remainder is \(-16\). Hence the result of the division is \(2x - 3 - \frac{16}{x-2}\).
04

Validate the Answer

To verify the quotient is correct, it is necessary to ensure that the divisor times the quotient, plus the remainder, equals the initial dividend, i.e., \((x - 2)*(2x - 3) - 16 = 2x^{2} + x - 10\).

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