What is (7(x^2+x-6))/((x^2-4x+4)(x^2-9)) ?

The solution to (7(x^2+x-6))/((x^2-4x+4)(x^2-9)) is 7/((x-2)(x-3))

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simplify (7(x^2+x-6))/((x^2-4x+4)(x^2-9))

simplify

\frac{7 ( x ^{2} + x - 6 )}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 9 )}

\frac{7}{( x - 2 ) ( x - 3 )}

Solution steps

\frac{7 ( x ^{2} + x - 6 )}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 9 )}

Factor

x^{2} + x - 6 : (x - 2) (x + 3)

= \frac{7 ( x - 2 ) ( x + 3 )}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 9 )}

Factor

x^{2} - 4 x + 4 : (x - 2)^{2}

= \frac{7 ( x - 2 ) ( x + 3 )}{( x - 2 ) ^{2} ( x ^{2} - 9 )}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(x - 2)^{2} = (x - 2) (x - 2)

= \frac{7 ( x - 2 ) ( x + 3 )}{( x - 2 ) ( x - 2 ) ( x ^{2} - 9 )}

Cancel the common factor:

x - 2

= \frac{7 ( x + 3 )}{( x - 2 ) ( x ^{2} - 9 )}

Factor

x^{2} - 9 : (x + 3) (x - 3)

= \frac{7 ( x + 3 )}{( x - 2 ) ( x + 3 ) ( x - 3 )}

Cancel the common factor:

x + 3

= \frac{7}{( x - 2 ) ( x - 3 )}

8 (- 3)

20 x^{4} y z^{6} + 12 x^{2} 5 y z^{6}

(2^{3} \cdot 5^{- 2}) (2^{- 2} \cdot 5^{4})

(3 x + 4 y) + (2 x - 2 y)

0 \cdot ln (0)

What is (7(x^2+x-6))/((x^2-4x+4)(x^2-9)) ?
The solution to (7(x^2+x-6))/((x^2-4x+4)(x^2-9)) is 7/((x-2)(x-3))