SOLUTION: Please help and show all work. Thanks Real Zeros of Polynomials 1. The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicit

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help and show all work. Thanks Real Zeros of Polynomials 1. The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicit      Log On


   

Question 1023621: Please help and show all work. Thanks


Real Zeros of Polynomials
1. 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=−2. It goes through the point (5,14).
Find a formula for P(x).
P(x)=

2. The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=5 and root of multiplicity 1 at x=-1. The y-intercept is y=-12.5.
Find a formula for P(x).
P(x)=
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1. A polynomial with a root of multiplicity 2 at x=3 has %28x-3%29%5E2 as a factor.
A polynomial with roots of multiplicity 1 at x=0 and x=−2
has %28x-0%29=x and %28x-%28-2%29%29=%28x%2B2%29 as factors.
The factors listed make
P%28x%29=a%2Ax%2A%28x%2B2%29%2A%28x-3%29%5E2 , with a%3C%3E0 a polynomial f degree 4.
If the graph of such polynomial goes through point (5,14), then P%285%29=14 ,and a%2A5%2A%285%2B2%29%2A%285-3%29%5E2=14 , so
a%2A5%2A7%2A2%5E2=14-->a%2A5%2A7%2A4=14-->a=14%2F%285%2A7%2A4%29-->a=0.1 .
So, highlight%28P%28x%29=0.1%2Ax%2A%28x%2B2%29%2A%28x-3%29%5E2%29 is a formula for such a polynomial.

2. Similarly, the polynomial must be of the form
P%28x%29=a%2A%28x%2B1%29%2A%28x-5%29%5E2 , with a%3C%3E0 ,
ana if the y-intercept is -12.5 ,
a must be such that P%280%29=12.5 .
So, a%2A%280%2B1%29%2A%280-5%29%5E2=12.5-->a%2A1%2A25=12.5-->a%2A25=12.5-->a=12.5%2F25-->a=0.5,
and highlight%28P%28x%29=0.5%2A%28x%2B1%29%2A%28x-5%29%5E2%29 is a formula for such a polynomial.