Question 548300
Keep this in mind.  It's taken from another answer, but describes the situation well.<P>
Consider:
(x-a)(x-b)(x-c)=0<P>
Recall from properties of numbers that if you multiply any quantity by 0 you get 0. So at least one of x-a, x-b, x-c has to be 0 for this to be true. What makes that true? x=a,x=b,x=c!<P>
Also remember FOIL when multiplying two binomials.<P>
a. 1,1 is x=1 and x=1 meaning (x-1)(x-1) or {{{(x-1)^2 = x^2 - 2x +1}}}<P>
b. 0,5,-5 is x=0, x=5 and x=-5 meaning x(x-5)(x+5) = {{{x(x^2-25)=x^3-25x}}}<P>
c.0,0,0,1 is x=0, x=0, x=0, x=1 meaning x*x*x(x-1) = {{{x^3 (x-1) = x^4 - x^3}}}
d. 2,2,1,-1 is x=2, x=2, x=1, x=-1 meaning {{{(x-2)^2 *(x-1)*(x+1)}}}
{{{(x^2 - 4x + 4)(x-1)(x+1)=(x^3-5x^2+8x-4)(x+1)=x^4-4x^3+3x^2+4x-4}}}
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