E ( x ) = ( \frac { x + 3 } { x + 1 } - \frac { x + 1 } { x + 3 }
Solve for E
E=\frac{4\left(x+2\right)}{x\left(x+1\right)\left(x+3\right)}
x\neq -3\text{ and }x\neq -1\text{ and }x\neq 0
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Ex\left(x+1\right)\left(x+3\right)=\left(x+3\right)\left(x+3\right)-\left(x+1\right)\left(x+1\right)
Multiply both sides of the equation by \left(x+1\right)\left(x+3\right), the least common multiple of x+1,x+3.
Ex\left(x+1\right)\left(x+3\right)=\left(x+3\right)^{2}-\left(x+1\right)\left(x+1\right)
Multiply x+3 and x+3 to get \left(x+3\right)^{2}.
Ex\left(x+1\right)\left(x+3\right)=\left(x+3\right)^{2}-\left(x+1\right)^{2}
Multiply x+1 and x+1 to get \left(x+1\right)^{2}.
\left(Ex^{2}+Ex\right)\left(x+3\right)=\left(x+3\right)^{2}-\left(x+1\right)^{2}
Use the distributive property to multiply Ex by x+1.
Ex^{3}+4Ex^{2}+3Ex=\left(x+3\right)^{2}-\left(x+1\right)^{2}
Use the distributive property to multiply Ex^{2}+Ex by x+3 and combine like terms.
Ex^{3}+4Ex^{2}+3Ex=x^{2}+6x+9-\left(x+1\right)^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.
Ex^{3}+4Ex^{2}+3Ex=x^{2}+6x+9-\left(x^{2}+2x+1\right)
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.
Ex^{3}+4Ex^{2}+3Ex=x^{2}+6x+9-x^{2}-2x-1
To find the opposite of x^{2}+2x+1, find the opposite of each term.
Ex^{3}+4Ex^{2}+3Ex=6x+9-2x-1
Combine x^{2} and -x^{2} to get 0.
Ex^{3}+4Ex^{2}+3Ex=4x+9-1
Combine 6x and -2x to get 4x.
Ex^{3}+4Ex^{2}+3Ex=4x+8
Subtract 1 from 9 to get 8.
\left(x^{3}+4x^{2}+3x\right)E=4x+8
Combine all terms containing E.
\frac{\left(x^{3}+4x^{2}+3x\right)E}{x^{3}+4x^{2}+3x}=\frac{4x+8}{x^{3}+4x^{2}+3x}
Divide both sides by x^{3}+4x^{2}+3x.
E=\frac{4x+8}{x^{3}+4x^{2}+3x}
Dividing by x^{3}+4x^{2}+3x undoes the multiplication by x^{3}+4x^{2}+3x.
E=\frac{4\left(x+2\right)}{x\left(x+1\right)\left(x+3\right)}
Divide 8+4x by x^{3}+4x^{2}+3x.
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