Question

Find the polynomial f(x) that has the roots of -2, 3 of multiplicity 2. Explain how you would verify the zeros of f(x).

Answer 1

hence you have x=-2,3

Answer 2

But the multiplicity is 2, so I would have (x+2)(x-3)(x-3) = 0

Answer 3

yeah, i forgot

Answer 4

you just need to expand that factors into a polynomial

Answer 5

so you'll have \[(x^2 -6x +9)(x+2)\]
right?

Answer 6

To verify your finished work, plug in the roots and make sure you get \(f(x) = 0\)

Answer 7

then distribute x+2
\[x^2(x+2)-6x(x+2)-9(x+2)\]

Answer 8

What I got was x^3 -6x^2 -19x - 18

Answer 9

check your work

Answer 10

x^3 - 4x^2 - 3x + 18

Answer 11

**-21x-18

Answer 12

\[\text{Factor}[x{}^{\wedge}3-4x{}^{\wedge}2-3x+18]\]\[(-3+x)^2 (2+x)\]

Answer 13

And I messed up my previous post. Sigh.

\[x^3-4x^2-3x+18\] is correct.