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Chapter 5: Problem 21

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{6 x^{2}-5 x-30}{2 x-5}$$

Short Answer

Expert verified
The result of the division is \(3x-1\) with no remainder. The check of the division is successful as the product of the divisor and the quotient, plus the remainder, equals the original polynomial (the dividend).

Step by step solution

01

Polynomial Division Setup

Arrange the polynomial \(6 x^{2}-5 x-30\) and \(2 x-5\) as a long division problem. Write \(6 x^{2}-5 x-30\) under the division symbol and \(2 x-5\) outside.
02

Perform the Polynomial Division

Divide the leading term of the dividend (\(6 x^{2}\)) by the leading term of the divisor (\(2x\)). This results in \(3x\) which is the first term of the quotient. Now multiply the divisor (\(2x-5\)) by the first term of the quotient (\(3x\)), and subtract the result from the dividend (\(6 x^{2}-5 x-30\)) . This gives us new dividend (\(0x -15x\)). Repeat the division step by dividing leading term of new dividend and leading term of divisor. This gives \( -1\) which is the second term of the quotient. Multiply this with divisor and subtract from new dividend to get the remainder of \(0\). Thus, the quotient is \(3x-1\) and remainder is \(0\).
03

Check the Result

To verify the result, multiply the quotient (\(3x-1\)) with divisor (\(2x-5\)), and add the remainder (\(0\)). The result should be equal to the original polynomial (the dividend). In this case, multiplying \(3x-1\) and \(2x-5\) indeed results in \(6 x^{2}-5 x-30\), which is our original polynomial. Thus, our answer is correct.

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