Question 2239
{{{(x-3)(x^2+3x-4)=0}}}
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Factor out the second equation that's in parentheses...
(x+4)(x-1)
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Now we get the 3 factors...
(x-3)(x+4)(x-1)=0
Now solve for x in each parentheses, and we get..
x=3, x=-4, x=1
<b>{3, -4, 1)</b>
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{{{(x+2)(x^2+5x+1)=0}}}
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The second equation cannot be factored and therefore, must be solved using the quadratic formula, which is {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
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Plug the values in, like this
{{{x = (-5 +- sqrt( 5^2-4*1*1))/(2*1)}}}
Which simplifies to...
{{{x=(-5+-sqrt(21))/2}}}
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So the answer to the second problem you gave is...
{-2,{{{(-5-sqrt(21))/2}}},{{{(-5-sqrt(21))/2}}}}
Hope this helps!
MS