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Chapter 5: Problem 18

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{2 b^{2}-9 b-5}{2 b+1}$$

Short Answer

Expert verified
The quotient of the given division \(\frac{2 b^{2}-9 b-5}{2 b+1}\) is \(b - 5\).

Step by step solution

01

Setting Up the Long Division

Set up the long division, similar to how traditional numeric long division is set up. Write the dividend \(2b^2 - 9b - 5\) inside the L-shaped symbol and the divisor \(2b + 1\) outside.
02

Divide

Divide the first term of the dividend by the first term of the divisor. In this case, \(2b^2\) divided by \(2b\) is \(b\). Write \(b\) above the division symbol as the first term of the quotient.
03

Multiply and Subtract

Multiply the divisor by the first term of the quotient, then subtract this result from the original dividend. Here, \(b\) times \(2b + 1\) gives \(2b^2 + b\). Subtracting this from \(2b^2 - 9b - 5\) gives \(-10b - 5\) as the new dividend.
04

Repeat the Division

Repeat the division with the new dividend. So, \(-10b\) divided by \(2b\) is \(-5\). Write \(-5\) above the line, next to \(b\). Now the quotient is \(b - 5\). Again, multiply \(-5\) times \(2b + 1\) obtaining \(-10b - 5\). Subtract this from \(-10b - 5\), resulting in 0, which means there is no remainder.
05

Checking the Result

To confirm the correctness of the result, multiply the divisor with the quotient and check that the result is the original dividend. Here that means calculate \((2b + 1) \times (b - 5)\) to confirm this equals \(2b^2 - 9b - 5\).
06

Simplify and Write the Final Answer

The polynomial division is now complete with quotient \(b - 5\) and no remainder. This is the final result.

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