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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\huge x^3-3x^2=4x-12\]
Rearrange as \[x^3-3x^2-4x+12 = 0\]Our factors will be of the form \[(x+a)(x+b)(x+c) = 0\] to generate the x^3 term, and because the 12 is positive, we know that either all of the values are positive, or two of them are negative, because those are the only combinations of three numbers that can be multiplied together and yield a positive number. However, because the middle terms of our polynomial are negative, we must have some negative values, so 2 are negative and 1 is positive. What are the prime factors of 12? A bit of experimentation will give you the combination that unlocks the factoring of the polynomial. Once you've factored the polynomial, you'll have (x+a) = 0, (x+b) = 0, (x+c) = 0 and the values of a, b, and c that make those expressions = 0 are the solutions or roots or zeros.

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