What is (b^2+3b-4)/(b^2+9b+20) ?

The solution to (b^2+3b-4)/(b^2+9b+20) is (b-1)/(b+5)

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simplify (b^2+3b-4)/(b^2+9b+20)

simplify

\frac{b ^{2} + 3 b - 4}{b ^{2} + 9 b + 20}

\frac{b - 1}{b + 5}

Solution steps

\frac{b ^{2} + 3 b - 4}{b ^{2} + 9 b + 20}

Factor

b^{2} + 3 b - 4 : (b - 1) (b + 4)

= \frac{( b - 1 ) ( b + 4 )}{b ^{2} + 9 b + 20}

Factor

b^{2} + 9 b + 20 : (b + 4) (b + 5)

= \frac{( b - 1 ) ( b + 4 )}{( b + 4 ) ( b + 5 )}

Cancel the common factor:

b + 4

= \frac{b - 1}{b + 5}

\frac{3 ^{- 3}}{6 ^{- 2}}

(1 + i) (1 - 2 i)

\frac{2}{3} + (- \frac{1}{4})

(- 6 m + 6) (3 m^{2} + 4 m - 3)

2 x (x + 1)^{3}

What is (b^2+3b-4)/(b^2+9b+20) ?
The solution to (b^2+3b-4)/(b^2+9b+20) is (b-1)/(b+5)