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Polynomial

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2 Answers

r1 = root 1 ---> x - r1 = 0

r2 = root 2 ----> x - r2 = 0

r3 = root 3 ----> x - r3 = 0

you just line them up and distribute

(x - r1)(x - r2)(x - r3) = 0

and you end up with

x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - (r1*r2*r3) = 0

 

Remember that the roots satisfy f(x)=0.  Some time ago you solved polynomials by factoring, and used the zero-product property to find the roots.  For example, suppose you factored x2-4x-5=0 and got (x+1)(x-5)=0.  By the zero-product property, your roots are -1 and 5.

Going from the roots to the polynomial is like going backwards.  Suppose you are given -2, 3, and 0 as roots.  You could write it in the factored form:

x(x+2)(x-3)=0

(You can check by plugging in the roots and seeing that the result=0.) 

Now to find the polynomial, multiply the terms.

x(x+2)(x-3) = (x2+2x)(x-3) = x3-3x2+2x2-6x = x3-x2-6x  
So in my sample, x3-x2-6x is the polynomial with roots -2, 0, and 3

I hope this answers your question!  Let me know if you need clarification.