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Chapter 5: Problem 15

Divide as indicated. Check each answer by showing that the product of the divisor and the quotient, plus the remainder, is the dividend. $$\frac{12 y^{2}-20 y+3}{2 y-3}$$

Short Answer

Expert verified
The given expression \(\frac{12 y^{2}-20 y+3}{2 y-3}\) simplifies to \(6y - 1\).

Step by step solution

01

Dividing First Terms

Firstly, divide the term with the highest power from the numerator, which is \(12y^{2}\), by the term with the highest power of the denominator, which is \(2y\). This will result in \(6y\), the first term of your quotient.
02

Subtracting Polynomials

Next, multiply the first term of the quotient, \(6y\), by the denominator, \(2y-3\), obtain \(12y^{2}-18y\). Subtract this polynomial from the original one, \(12y^{2}-20y+3 - (12y^{2}-18y)\), result in \(-2y + 3\).
03

Dividing First Terms Again

Now divide the term with the highest power from the new polynomial and the term with the highest power from the denominator. This gives us \(\frac{-2y}{2y} = -1\) which is the second term of our quotient.
04

Subtracting Polynomials Again

Multiply the second term of quotient by the denominator and subtract from the new polynomial. We will obtain our remainder: \(-2y + 3 - ((-1) \cdot (2y - 3)) = 0\). In this case, our remainder is 0, which means the original polynomial was completely divisible by the denominator.
05

Combining Terms

Lastly, combine the terms we have obtained, which were \(6y\) and \(-1\), to give us our final quotient of \(6y - 1\).

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\left(4 \times 10^{3}\right)+\left(3 \times 10^{2}\right)=4.3 \times 10^{3}$$

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. If \(5^{-2}\) is raised to the third power, the result is a number between 0 and 1

In each exercise, find the product. $$(x+3)\left(x^{2}+5\right)$$

Simplify: \(24+8 \cdot 3+28 \div(-7)\).

In Exercises \(105-106,\) find the missing coefficients and exponents designated by question marks. $$\frac{3 x^{14}-6 x^{12}-7 x^{7}}{2 x^{7}}=-x^{7}+2 x^{5}+3$$
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