# Please help me solve the following math problems: `((t^2+3t-18)/(t^2+2t-15))((t^2-3t-10)/(t^2+8t+12))` (Only one question, and one related follow up question, please)

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### 1 Answer

The first step is to factor the 4 polynomials. Find two numbers that multiply to the last term, and add to the constant of the second term:

`t^2+3t-18=(t+6)(t-3)`

6+-3=3 and (6)(-3)=-18

`t^2-3t-10=(t-5)(t+2)`

-5+2=-3 and (-5)(2)=10

`t^2+2t-15=(t+5)(t-3)`

5+-3=2 and (5)(-3)=-15

`t^2+8t+12=(t+6)(t+2)`

6+2=8 and (6)(2)=12

Therefore:

`((t^2+3t-18)/(t^2+2t-15))((t^2-3t-10)/(t^2+8t+12))=(((t+6)(t-3))/((t+5)(t-3)))(((t-5)(t+2))/((t+6)(t+12))) `

Cancel out the equivalent terms, and you are left with:

`((t-5)(t+2))/((t+5)(t+12))`

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