recall that for any root say x - a, namely x - a = 0, thus x = a, for such a root of a polynomial, f(x), based on the remainder theorem f(a) = 0.
to put it in different lingo, since we know that we have two factors (x - 2 ) and (x+1), that simply means we have two roots of x = 2 and x = -1, well, using the remainder theorem that simply means f(2) = 0 and f(-1) = 0.
[tex] \bf x^3+ax^2+2x+b\\\\
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\stackrel{x=2}{f(2)}=(2)^3+a(2)^2+2(2)+b\implies \stackrel{\stackrel{f(2)}{\downarrow }}{0}=8+4a+4+b
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\boxed{0=4a+b+12}
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\stackrel{x=-1}{f(-1)}=(-1)^3+a(-1)^2+2(-1)+b\implies \stackrel{\stackrel{f(x)}{\downarrow }}{0}=-1+a-2+b
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0=a+b-3\implies \boxed{3-b=a}\\\\
------------------------------- [/tex]
[tex] \bf \stackrel{\textit{\textit{using the a value in the first equation found}}}{0=4(3-b)+b+12}\implies 0=12-4b+b+12
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0=24-3b\implies 3b=24\implies b=\cfrac{24}{3}\implies b=8\\\\
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\stackrel{\textit{and since we know that }}{3-b=a}\implies 3-8=a\implies -5=a\\\\
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x^3-5x^2+2x+8 [/tex]
and you can graph that, you'll see the roots or x-intercepts of -1 and 2 there.
