OpenStudy (anonymous):

(x+2)(x+3)(x+4)

4 years ago
OpenStudy (whpalmer4):

Are you to multiply it out, or what?

4 years ago
OpenStudy (lncognlto):

Use the FOIL method on the first two, and then multiply the quadratic by the remaining x term would be my approach.

4 years ago
OpenStudy (anonymous):

Yes

4 years ago
OpenStudy (whpalmer4):

just say no to FOIL, and learn the distributive property instead, which works for any number of terms, unlike FOIL. \[(x+a)(x+b)(x+c) = (x+a)(x(x+c) + b(x+c)) \]\[\qquad= (x+a)(x*x + c*x + b*x + b*c)\]\[\qquad=(x+a)(x^2+cx+bx+bc) = x(x^2+cx+bx+bc) + a(x^2+cx+bx+bc)\]\[\qquad =x^3+cx^2+bx^2+bcx+ax^2+acx+abx + abc\]\[\qquad=x^3+(a+b+c)x^2+(ab+ac+bc)x + abc\]

4 years ago
OpenStudy (whpalmer4):

sorry, should have split that middle line in two so that OpenStudy would show all of it, but I think you get the idea.

4 years ago
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