SOLUTION: The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=−3 Find a possible formula for P(x).

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=−3 Find a possible formula for P(x).       Log On


   

Question 1200280: The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=−3
Find a possible formula for P(x).
Found 3 solutions by ikleyn, josgarithmetic, math_tutor2020:
Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.

    P(x) = x%5E2%2A%28x-1%29%5E2%2A%28x%2B3%29.    ANSWER



Answer by josgarithmetic(39613) About Me  (Show Source):
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: P%28x%29+=+x%5E2%28x-1%29%5E2%28x%2B3%29


Explanation:
If k is a root of P(x), then x-k is a factor of P(x)
The multiplicity is the exponent for the factor.
So that's how for instance a root of x = 1 with multiplicity 2 leads to the factor %28x-1%29%5E2

The leading coefficient is the number out front of the variables.
For example, if the leading coefficient was 5, then we would have P%28x%29+=+5x%5E2%28x-1%29%5E2%28x%2B3%29

Instead, the leading coefficient is 1 so we have P%28x%29+=+1x%5E2%28x-1%29%5E2%28x%2B3%29 aka P%28x%29+=+x%5E2%28x-1%29%5E2%28x%2B3%29

Optionally you can expand everything out to get P%28x%29+=+x%5E2%28x-1%29%5E2%28x%2B3%29+=+x%5E5%2Bx%5E4-5x%5E3%2B3x%5E2
However, I don't recommend doing this since it's unnecessary and you lose the information about the roots along with their multiplicities.