Let's say y1, y2, and y3 are zeros. Set up an expression like this (x-y1)(x-y2)(x-y3) [This is factored form] and then multiply carefully. That works with any number of roots. If 0 is a root, add an x by itself to the beginning of the factored form expression. Also, imaginary roots come in twos. Therefore, if given i as a root, you need (x-i)(x+i). If you have something in x + yi form for a root, you'll need the complex conjugate. That would make (x-(a+bi))(x-(a-bi))
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If the cubic polynomial you are given does not have an obvious factorization, then you must use synthetic division. I'm sure wikipedia can tell you all about that.
You solve the equation.
(1,2) (0,5) (-1,8) (2,-1) (-2,11) All of these are solutions to the given equation.
Ah, don't you worry, friend. In a Mega Millions jackpot, there are quite a few zeros! You'll find six zeros in a million and nine zeros in a billion. Just imagine all the happy little zeros lining up to bring joy and excitement to someone's life.
You cannot graph quadratics by finding its zeros: you need a lot more points.Some quadratics will have no zeros. Having the zeros does not tell you whether the quadratic is open at the top (cup or smiley face) or open at the bottom (cap or grumpy face). Furthermore, it gives no indication as to how far above, or below, the apex is.