SOLUTION: The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicity 1 at x=0 and x=−3. It goes through the point (5,112). Find a formula for P(x).

Algebra ->  Equations -> SOLUTION: The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicity 1 at x=0 and x=−3. It goes through the point (5,112). Find a formula for P(x).       Log On


   

Question 1132254: The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicity 1 at x=0 and x=−3. It goes through the point (5,112).
Find a formula for P(x).
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The polynomial of degree+4,
P%28x%29=%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29%28x-x%5B4%5D%29

P(x) has a root of multiplicity 2 at x=3+
x%5B1%5D=x%5B2%5D=3

and root of multiplicity 1 at x=0+
x%5B3%5D=0
and root of multiplicity 1 atx=-3
x%5B4%5D=-3
P%28x%29=%28x-x%5B1%5D%29%5E2%28x%29%28x-x%5B4%5D%29
P%28x%29=%28x-3%29%5E2%28x%29%28x%2B3%29
It goes through the point (5,112).
P%28x%29=a%28x-3%29%5E2%28x%29%28x%2B3%29
P%285%29=a%28x-3%29%5E2%28x%29%28x%2B3%29
P%285%29=a%285-3%29%5E2%285%29%285%2B3%29
P%285%29=a%282%29%5E2%285%29%288%29
P%285%29=4a%2840%29
P%285%29=160a
since P%285%29=112=>160a=112=>a=112%2F160=>a=7%2F10
P%28x%29=%287%2F10%29%28x-3%29%5E2%28x%29%28x%2B3%29
P%28x%29=%287%2F10%29%28x%5E4+-+3+x%5E3+-+9+x%5E2+%2B+27+x%29