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Algebra
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How do I factor: x^3-7x-6
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I'm sorry if it is wrong... I am a bit rusty at this but this is the best I could do. x^3 - 7x - 6 = x^3 - 2x^2 - 3x + 2x^2 - 4x - 6 = (x^3 - 2x^2 - 3x) + (2x^2 - 4x - 6) = x(x^2 - 2x - 3) + 2(x^2 - 2x - 3) = (x^2 - 2x - 3)(x + 2) = (x^2 + x - 3x - 3)(x + 2) = [(x^2 + x) - (3x + 3)](x + 2) = [x(x + 1) - 3(x + 1)](x + 2)
That's alright. It looks right, since I'm understanding how you did it. I'll try it out using your steps to see if I get the same answer. Thank you!
No problem!
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