I'm trying to solve x^3-3x^2-x+3 is less than zero. Except, I'm not sure how to factor it. I tried splitting it so that 3 would be a factor for 3x^2 and 3, so the result was 3(-x^2+1), and x^2 was a factor for x^3 and -3x^2, resulting in x^2(x+3). But this will be really hard to find out the zeroes from. Can you explain a better way? Thanks!

asked by Devon

15 years ago

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1 answer

x^3-3x^2-x+3
x^2(x-3)-(x-3)
(x-3)(x^2-1)
(x-3)(x+1)(x-1)

answered by bobpursley

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I'm trying to solve x^3-3x^2-x+3 is less than zero. Except, I'm not sure how to factor it. I tried splitting it so that 3 would be a factor for 3x^2 and 3, so the result was 3(-x^2+1), and x^2 was a factor for x^3 and -3x^2, resulting in x^2(x+3). But this will be really hard to find out the zeroes from. Can you explain a better way? Thanks!

1 answer

x^3-3x^2-x+3
x^2(x-3)-(x-3)
(x-3)(x^2-1)
(x-3)(x+1)(x-1)

Ask a New Question or answer this question.

bobpursley · Answer

x^3-3x^2-x+3 x^2(x-3)-(x-3) (x-3)(x^2-1) (x-3)(x+1)(x-1)