The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=5 and a root of multiplicity 1 at x=−5. The y-intercept is y=−87.5. Find a formula for P(x). P(x)=

Answer 1

P(x) = (x-a)³+(x-b)²+(x-c)+d ----? general rule of thumb if you have a third power polynomial then you need 4 parts of information, if you have a 2 power polynomial you need 3 parts of information...

when x=0, y= -87.5 (y-intercept) = d

root#1: a=0 (no information given) With the information given, there is no 4th term with a cubic power, or multiplicity

root #2, b=5, power/multiplicity = 2: (x-5) with multiplicity of two : (x-5)²

root#3, c= -5, power/multiplicity = 1 : (x--5) ----> (x+5)¹ = (x+5)

Rewrite our equation:

P(x) = (x-5)²+(x+5)-87.5

With the information given, there is no 4th term with a cubic power, or multiplicity

Answer 2

Hi Mikayla,

There's a relationship between the roots of a polynomial and the factors of the polynomial. In particular, if r is a root of multiplicity 1, then (x-r) must be a factor. If s is a root of multiplicity 2, then (x-s)² must also be a factor. Since the polynomial is degree 3 there are no other factors, so P(x) must equal some constant k times (x-r) times (x-s)². Use the fact that the y-intercept is equal to -87.5 to determine k.

Answer 3

A root of a polynomial is the x value that makes y=0. If there is a root at x=5, then the polynomial must have a factor (x-5) [Set x to 5, this term will evaluate to 0, since it is a factor of the polynomial, the whole expression will be 0)

The multiplicity of this root is 2, which means there are two terms like this:

P(x) = (x-5)^2...

The other root is at -5, so we have a term (x - (-5)) = (x + 5). Putting all these together:

P(x) = (x-5)^2 * (x+5) ...

But, there may be a constant multiplier, let's call it A. So, we have

P(x) = A * (x-5)^2 * (x+5)

y-intercept means, the value of y when x is 0. When we set x = 0, y will be -87.5

-87.5 = A (0-5)^2 * (0+5)

-87.5 = A * 125

A = -87.5 / 125