What is the answer to (((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0 ?

The answer to (((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0 is x=1,x=-2

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(((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0

\frac{( ( 1 - x ) ^{3} ( x + 2 ) ^{4} )}{( ( x + 9 ) ^{2} ( x - 8 ) )} = 0

x = 1, x = - 2

Solution steps

\frac{( ( 1 - x ) ^{3} ( x + 2 ) ^{4} )}{( ( x + 9 ) ^{2} ( x - 8 ) )} = 0

Simplify

\frac{( ( 1 - x ) ^{3} ( x + 2 ) ^{4} )}{( ( x + 9 ) ^{2} ( x - 8 ) )} : \frac{( 1 - x ) ^{3} ( x + 2 ) ^{4}}{( x + 9 ) ^{2} ( x - 8 )}

\frac{( 1 - x ) ^{3} ( x + 2 ) ^{4}}{( x + 9 ) ^{2} ( x - 8 )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

(1 - x)^{3} (x + 2)^{4} = 0

Solve

(1 - x)^{3} (x + 2)^{4} = 0 : x = 1, x = - 2

x = 1, x = - 2

Verify Solutions

Find undefined (singularity) points:

x = - 9, x = 8

Combine undefined points with solutions:

x = 1, x = - 2

9.6 = \frac{- 1}{10 ( x - 12 ) ^{2} + 10}

2 (\frac{x - 1}{x})^{2} - 5 (\frac{x + 1}{x + 2}) + 18 = 0

\frac{( x ^{- 2} )}{x} = \frac{1}{( x ^{12} )}

\frac{( x - 6 )}{( x + 4 )} = 0

2 x^{- 5} = (16)^{2}

What is the answer to (((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0 ?
The answer to (((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0 is x=1,x=-2
Solve for x: (((1-x)^3(x+2)^4))/(((x+9)^2(x-8)))=0
The solution is x=1,x=-2