Problem 15 - Introductory Algebra for College Students

Question

Multiply as indicated. $$\frac{y^{2}-7 y-30}{y^{2}-6 y-40} \cdot \frac{2 y^{2}+5 y+2}{2 y^{2}+7 y+3}$$

Step-by-Step Solution

Answer

The result of the multiplication is \(\frac{y^{3}+5y^{2}-19y-43}{y^{3}+7y^{2}-28y-81}\)

Step 1: Set up the problem

First, set up the problem with the two fractions that need to be multiplied. This is done by multiplying the numerators together and the denominators together. So, we have \((y^{2}-7y-30)\cdot (2y^{2}+5y+2)\) over \((y^{2}-6y-40)\cdot (2y^{2} +7y+3)\)

Step 2: Multiply the numerators

Multiply the numerators together. This should produce the polynomial: \(2y^{4}+5y^{3}-19y^{2}-86y-60\)

Step 3: Multiply the denominators

Multiply the denominators together. This results in the polynomial: \(2y^{4}+7y^{3}-28y^{2}-162y-120\)

Step 4: Insert the results into the fraction

Place the result from the numerators and denominators multiplication into the fraction. It should be \(\frac{2y^{4}+5y^{3}-19y^{2}-86y-60}{2y^{4}+7y^{3}-28y^{2}-162y-120}\)

Step 5: Cancel common terms

In this fraction, \(2y^4\) is common in the numerator and the denominator. Also, the constants -60 and -120 are common. It can be reduced to \(\frac{y^{3}+5y^{2}-19y-43}{y^{3}+7y^{2}-28y-81}\). This is the simplest form of the fraction that can be obtained.