Finding the Formula for a Polynomial

Question

The polynomial of degree , , has a root of multiplicity  at  and a root of multiplicity  at . The -intercept is .

Answer 1

The Fundamental Theorem of Algebra and the given problem conditions imply that

P(x)=a_{3}*(x-3)^2*(x-(-3))=a_{3}*(x-3)^2*(x+3)

and

P(0)=-2.7.

Since we find that

P(0)=a_{3}*(0-3)^2*(0+3)=a_{3}*(-3)^2*3=27*a_{3}=-2.7,

we can also deduce that

a_{3}=-2.7/27=-0.1.

From all of the preceding, we can conclude that

P(x)=-0.1*(x-3)^2*(x+3)

=-0.1*(x^2-2*3*x+3^2)*(x+3) [since (x-a)^2=x^2-2ax+a^2]

=-0.1*(x^2-6x+9)*(x+3)

=-0.1*(x^3-6x^2+9x+3x^2-18x+27)

=-0.1*(x^3-3x^2-9x+27)

=-0.1x^3+0.3x^2+0.9x-2.7