How do you simplify and divide $(g^{2} + 8 g + 15) {(g + 3)}^{- 1}$ $(g^2+8g+15)(g+3)^-1$ ?

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Answer 1 · 2016-12-05T11:12:16

$\frac{g^{2} + 8 g + 15}{g + 3} = ? ?$ $(g^2 + 8g+15)/(g+3)=??$

Factorize the numerator..........

And thus, $g^{2} + 8 g + 15 = (g + 5) (g + 3)$ $g^2+8g+15=(g+5)(g+3)$

And thus, $(g^2 + 8g+15)/(g+3)=[(g+5)(g+3)]/(g+3)=[(g+5)cancel((g+3))]/cancel((g+3))$

$=$ $g+5$

How did I know how to factorize $(g^2 + 8g+15)$ ? Given the question, I knew that $(g+3)$ was likely to be a factor, and I proceeded from that basis.

How do you simplify and divide (g^2+8g+15)(g+3)^-1(g2+8g+15)(g+3)−1(g^2+8g+15)(g+3)^-1?