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find one of the factors when (a+3)^2 - 3(a+3)-18 is factored completely? a.a+6 b.a+3 c.a-6 d.a-2
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a^2 + 9 - 3a - 9 - 18 solve :)
?
Doing a slight change of variable, if we let x = a + 3, then the equation becomes;\[x^2-3x-18\]Now we need to think of two numbers that multiply to -18, and add up to -3.
-6 , 3?
Right :) so we now have:\[x^2-3x-18=(x-6)(x+3)\], but originally we had x = a+3, so putting that back in gives:\[(x-6)(x+3)=(a+3-6)(a+3+3)=(a-3)(a+6)\]
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Note: we really didnt have to do the variable substitution. Whenever you have things like:\[a(junk)^2+b(junk)+c=0\], and whatever is inside the parenthesis is the same, you can always treat it like:\[ax^2+bx+c=0\], factor it, and replace x with the junk.
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