SOLUTION: The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-5. find a possible formula for P(x)

Algebra ->  Expressions-with-variables -> SOLUTION: The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-5. find a possible formula for P(x)      Log On


   

Question 1183867: The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-5. find a possible formula for P(x)
Found 2 solutions by MathLover1, Solver92311:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29 is of degree 5
P%28x%29=a%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29%28x-x%5B4%5D%29%28x-x%5B5%5D%29
given:
-a leading coefficient 1: a=1
has roots of multiplicity 2+at+x=5: %28x-5%29%28x-5%29
has root of multiplicity 2 at x=0:%28x-0%29%28x-0%29
a root of multiplicity 1 at x=-5:%28x%2B5%29

P%28x%29=1%28x-5%29%28x-5%29%28x-0%29%28x-0%29%28x%2B5%29
P%28x%29=%28x-5%29%28x-5%29%28x%29%28x%29%28x%2B5%29
P%28x%29=x%5E5-5x%5E4-25x%5E3%2B125x%5E2

+graph%28+600%2C+600%2C+-7%2C+7%2C+-30%2C+1400%2Cx%5E5-5x%5E4-25x%5E3%2B125x%5E2%29+




Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!


If is a zero of a polynomial function, then is a factor of the polynomial. The number of factors of the polynomial is equal to the degree of the polynomial. So your polynomial has the following factors:









, which is to say

The set of all 5th-degree polynomials with real coefficients that have this set of zeros can be described as:



Where is the lead coefficient. Your lead coefficient is given to be 1, so:



is a valid possible formula for . You could expand that, but that is a lot of extra work that is not required by the question posed.

John

My calculator said it, I believe it, that settles it

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