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I am trying to solve the equation $ax^3+bx^2+(c-j)x+(d-m)=0$ to find an expression for all three roots. I found this 1 but whenever I try and solve it by substituting values in I always get imaginary numbers as answers for example the equation $x^3-2x+0.25$ I get imaginary numbers as the answers for x even when there are 3 clear real roots. I was wondering if somebody could run through a step by step solution to $x^3-2x+0.25$ or suggest another method for solving $ax^3+bx^2+(c-j)x+(d-m)=0$ ? Thanks

fosho
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Alex Jones
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1 Answers1

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Are you familiar with the cubic formula? Here's a link: http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Alternatively, in high school, for such polynomial equations, we tried substituting possible factors into the equation (e.g. for this case 1, 1/4, 1/2), then doing synthetic division on that.

Here's a more detailed note on factoring: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm

JM1
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  • Please do not have most of your answer rely on links, as linnks may be broken later. Instead, put what you want to say into your answer instead of linking. – Teoc Jan 03 '16 at 17:37
  • I tried with the cubic formula, however each time I did that I seemed to be getting a negative inside the square root and therefore an imaginary number even though it was clear there were real roots. And the substitution method doesn't work as I am trying to find a general formula – Alex Jones Jan 03 '16 at 18:42