Problem 58 - Introductory Algebra for College Students

Question

Divide as indicated. $$\frac{3 y+12}{y^{2}+3 y} \div \frac{y^{2}+y-12}{9 y-y^{3}}$$

Step-by-Step Solution

Answer

\[\frac{-3 y^{3}-12 y^{2}+27 y+108}{y^{3}+4y^{2}-21y-36}\]

Step 1: Rewrite the Division as Multiplication

By the 'keep, change, flip' rule, \[\frac{3 y+12}{y^{2}+3 y} \div \frac{y^{2}+y-12}{9 y-y^{3}} \]can be rewritten as\[\frac{3 y+12}{y^{2}+3 y} \cdot \frac{9 y-y^{3}}{y^{2}+y-12}\]

Step 2: Multiply the Fractions

We then multiply the numerators together and the denominators together to form the new fraction:\[\frac{(3 y+12)(9 y-y^{3})}{(y^{2}+3 y)(y^{2}+y-12)}\]

Step 3: Simplify the Fractions

Now, simplify the expression and factor where possible. The numerator becomes: \[27 y^{2}-3 y^{4}+108 y-12 y^{3}\]Rearrange them to match the standard polynomial format \(a x^{n}+b x^{n-1}+\ldots+k\):\[ -3 y^{4}-12 y^{3}+27 y^{2}+108 y\]The denominator becomes:\[y^{4}+4y^{3}+3y^{2}-12y^{2}-36y-12y^{2}\]Combine the like terms\[y^{4}+4y^{3}-21y^{2}-36y\]

Step 4: Find the Greatest commom factor (GCF)

From the numerator, we can factor out y to get\[ y(-3 y^{3}-12 y^{2}+27 y+108)\]From the denominator, we can factor out y too to get\[ y(y^{3}+4y^{2}-21y-36)\]Here the greatest common factor in both numerator and denominator is y. After cancelling it out, the expression becomes\[\frac{-3 y^{3}-12 y^{2}+27 y+108}{y^{3}+4y^{2}-21y-36}\]

Step 5: Final Simplification

The above expression is the final simplified result. It cannot be simplified any further.