SOLUTION: Find a degree 3 polynomial that has zeros -2, 3 and 6 and in which the coefficient of x^2 is -14? ok.. I know the zeros would be (x+2) (x-3) (x-6).. then I get lost with the -14

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find a degree 3 polynomial that has zeros -2, 3 and 6 and in which the coefficient of x^2 is -14? ok.. I know the zeros would be (x+2) (x-3) (x-6).. then I get lost with the -14      Log On


   

Question 975001: Find a degree 3 polynomial that has zeros -2, 3 and 6 and in which the coefficient of x^2 is -14?
ok.. I know the zeros would be (x+2) (x-3) (x-6).. then I get lost with the -14x^2..
can someone please help find what the polynomial is?
Found 2 solutions by josgarithmetic, jim_thompson5910:
Answer by josgarithmetic(39631) About Me  (Show Source):
You can put this solution on YOUR website!
What? Just finish! The binomial product makes the zeros work. That product by itself has leading coefficient of 1. Just apply the factor, -14.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Any cubic polynomial with roots -2,3,6 will be of this form

k%28x%2B2%29%28x-3%29%28x-6%29

where k is some constant

Let's expand out the polynomial


k%28x%2B2%29%28x-3%29%28x-6%29


k%28x%2B2%29%28x%5E2-9x%2B18%29


k%28x%28x%5E2-9x%2B18%29%2B2%28x%5E2-9x%2B18%29%29


k%28x%5E3-9x%5E2%2B18x%2B2x%5E2-18x%2B36%29


k%28x%5E3-7x%5E2%2B36%29


k%2Ax%5E3-7kx%5E2%2B36k


The x^2 term is -7kx%5E2


We want the x^2 coefficient to be -14, so, -7k=-14 which means k+=+2


Since k = 2, k%2Ax%5E3-7kx%5E2%2B36k turns into 2x%5E3-14x%5E2%2B72 after k is replaced with 2 and you simplify

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