SOLUTION: could you help me factor completely? i tried where am i wrong? x^2+2x-80 x^2+160-3x-80 (x^2+160x)-(3x+80) x^2(x+160)-3(x+80)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: could you help me factor completely? i tried where am i wrong? x^2+2x-80 x^2+160-3x-80 (x^2+160x)-(3x+80) x^2(x+160)-3(x+80)      Log On


   



Question 75538: could you help me factor completely? i tried where am i wrong?
x^2+2x-80
x^2+160-3x-80
(x^2+160x)-(3x+80)
x^2(x+160)-3(x+80)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+2x-80
Think of two numbers whose product is -80 and whose sum is +2.
They are +10 and -8
Rewrite the problem as:
x^2+10x-8x-80
Factor the 1st two terms and the last two terms separately:
=x(x+10)-8(x+10)
=(x+10)(x-8)
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This is called the AC Method of factoring.
Cheers,
Stan H.