Question 166405
Given:  {{{x^3+4x^2-9x+d=0}}}
Find d such that two zeros of the cubic are additive inverses {sum of the roots of a cubic ax³+bx²+cx+d is –b/a}. No guess and checking—support your answers algebraically!
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Ok, we know that we are going to factor this to find the zeros.  So let's think about how we want this to factor.  We want to be able to have two parenthesis that look identical (you will see what I am talking about in the problem).
We are going to group the first two terms and group the second two terms.
It now looks like this: {{{(x^3+4x^2)(-9x+d)=0}}}
we know that we are going to factor out {{{x^2}}} from the first set and we get
{{{x^2(x+4)+(-9x+d)=0}}}
what number "d" will allow us to factor out a number such that we are left with (x+4)?
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Well, we know that in order to make the first term = x, we need to factor out a -9 from both terms and the second term needs to end up a positive 4.  
So multiply -9 and 4 and you get d=-36.
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Now you have: {{{x^3+4x^2-9x+(-36)=0}}} or {{{x^3+4x^2-9x-36=0}}}
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Check to see if this indeed works by factoring everything and finding your zeroes.
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{{{(x^3+4x^2)(-9x-36)=0}}}
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{{{x^2(x+4)-9(x+4)=0}}}
factor out (x+4) from each term and you get
{{{(x+4)(x^2-9)=0}}}
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{{{x=-4}}}
{{{x=3}}}
{{{x=-3}}}
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