Respuesta :

[tex]\frac{4x^2+8x}{x^2-x-6}\\\\Domain:x^2-x-6\neq0\\\\x^2-3x+2x-3\cdot2=x(x-3)+2(x-3)=(x-3)(x+2)\\\\x^2-x-6\neq0\iff(x-3)(x+2)\neq0\iff x-3\neq0\ or\ x+2\neq0\\\\therefore\ x\neq3\ and\ x\neq-2\\------------------\\the\ nominator:\ 4x^2+8x=4x\cdot x+4x\cdot2=4x(x+2)\\the\ denominator:x^2-x-6=(x-3)(x+2)[/tex]

[tex]therefore\ simplify\ form:\\\\\frac{4x^2+8x}{x^2-x-6}=\frac{4x(x+2)}{(x-3)(x+2)}=\boxed{\frac{4x}{x-3}}[/tex]

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