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Chapter 5: Problem 28

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{4 y^{3}+3 y+5}{2 y-3}$$

Short Answer

Expert verified
The solution of the given polynomial division is \(2y^{2} + 3y + 2\).

Step by step solution

01

Set up the division

To begin, write out the division in long-division form. Here, \(4y^{3}+3y+5\) is the dividend (the number being divided) and \(2y-3\) is the divisor (the number we are dividing by). It's essentially the same as regular division, but with polynomials.
02

Divide

To divide a cubic polynomial by a linear one, you will divide the first term of the dividend by the first term of the divisor to find the first term of the quotient. This gives \((4y^{3} / 2y) = 2y^{2}\) . Multiply the divisor \((2y-3)\) by the first term of the quotient to get \(4y^{3}-6y^{2}\), then subtract this from the original dividend to get the new dividend \(6y^{2} + 3y + 5\). Repeat the process to continue the division.
03

Continue the division process

Now you repeat the division process with the new dividend and the original divisor until you can no longer continue the division process with the divisor. Using the same division process as before, we divide the first term of the new dividend by the first term of the divisor to get \(3y\). We now subtract this times the divisor from the new dividend to get \(9y+5\). We repeat this process one more time to get a final remainder of \(17\). So our final answer is \(2y^{2} + 3y + 2\).
04

Verify the result

Now, to confirm if your quotient \(2y^{2} + 3y + 2\) is correct, multiply it by the divisor \(2y-3\). The result should be the original dividend \(4y^{3}+3y+5\) after factoring out any common factors.
05

Write the final result

The final result of the polynomial division is \(2y^{2} + 3y + 2\). Thus dividing \(4y^{3}+3y+5\) by \(2y-3\) gives \(2y^{2} + 3y + 2\). Note that there is no remainder.

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