When doing sum and product of 0, -3, 2 the resulting quadratic is y=x^3+x^2-13x. The problem is that when we graphed it, it didn't pass through the zeros. Is there any reason for this? Nobody in my class or the teacher could figure out why. Does sum and product have restrictions or is there something else going on.
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so the 3 roots are $0,-3,2$? – The Integrator May 18 '18 at 18:26
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1While the roots of the cubic do have a sum and product that can be seen "by inspection" from the coefficients of the polynomial, you don't have enough "data" just from sum and product to determine all the coefficients. In other words, many cubic polynomials will have the same sum and product of three roots (counted according to their multiplicity). Try multiplying out $(x-0)(x+3)(x-2)$ to see this more clearly. – hardmath May 18 '18 at 18:28
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(a) What exactly do you mean by "doing sum and product"? Sorry, but unless you tell us exactly what you did, this doesn't make much sense. (b) Related to that: where did $x^3+x^2-13x$ come from? (c) The resulting polynomial is cubic, not "quadratic", because it's a degree $3$ polynomial. – zipirovich May 18 '18 at 19:05
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Zeros are at $\displaystyle 0$ and $\displaystyle{-1 \pm ,\sqrt{, 53,}, \over 2}$. Note that $\displaystyle \left(x - 0\right) \left[x - \left(-3\right)\right]\left(x - 2\right) = x^{3} + x^{2} - \color{red}{6}x$. – Felix Marin May 18 '18 at 19:13
