How do you simplify and divide $\frac{9 b^{2} + 9 b - 10}{3 b - 2}$ $(9b^2+9b-10)/(3b-2)$ ?

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How do you simplify and divide $\frac{9 b^{2} + 9 b - 10}{3 b - 2}$ $(9b^2+9b-10)/(3b-2)$ ?

1 Answer

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Answer 1 · 2016-10-16T08:36:47

$3 b + 5$ $3b+5$

Explanation:

Start by factorising the numerator
$9 b^{2} + 9 b - 10 = (3 b - 2) (3 b + 5)$ $9b^2+9b-10=(3b-2)(3b+5)$
Then $\frac{9 b^{2} + 9 b - 10}{3 b - 2} = \frac{(3 b - 2) (3 b + 5)}{3 b - 2} = 3 b + 5$ $(9b^2+9b-10)/(3b-2)=((3b-2)(3b+5))/(3b-2)=3b+5$
You can also do a long division and the result is the same

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How do you simplify and divide $\frac{9 b^{2} + 9 b - 10}{3 b - 2}$ $(9b^2+9b-10)/(3b-2)$ ?

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Explanation:

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How do you simplify and divide $\frac{9 b^{2} + 9 b - 10}{3 b - 2}$ $(9b^2+9b-10)/(3b-2)$ ?