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Please show me step by step how to factor this
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\[x ^{3}-3x ^{2}-4x+12\]
1, -1, 2, -2, 3 ,-3, 4, -4, 6, -6, 12 and -12 are all factors of 12. Do check and guess by plugging in each factor until you have one that equals zero. In this case, if you plug 2 into this equation, you get... \[(2)^3-3(2)^2-4(2)+12=8-12-8+12=0\]
That means x=2, or x-2=0
(x -3)(x + 2)(x - 2) This is the answer..i need to know how I ca arrive here
Then do... \[x^3-3x^2-4x+12 \div x-2\]
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and from there you'll have a polynomial that is easily factored
oooh...ok. I get what you are saying.
thanks!
No problem!
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