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multiply the polynomial below:
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\[\frac{ x^2-2x-8 }{ 6x^2-24x } \times \frac{ 2x^2-3x-2}{ x^2-4 }\]
i imagine your job is to factor as much as you can and then cancel
ok hold on let me try
for example \(x^2-2x-8=(x-4)(x+2)\) and you will also get a factor of \((x+2)\) in the denominator because \(x^2-4=(x+2)(x-2)\)
\[\frac{ 2x^2-7x-4 }{ x^2-4}\]
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i think you can do more
-7x?
-7x-1?
ok lets go ahead and factor
wait 2-7x +1 3-7x
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\[\frac{ x^2-2x-8 }{ 6x^2-24x } \times \frac{ 2x^2-3x-2}{ x^2-4 }\] \[\frac{(x-4)(x+2)(x-2)(2x+1)}{6x(x-4)(x+2)(x-2)}\]
cancel away and you are not left with much just \[\frac{2x+1}{6x}\]
thank you
yw
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