Question

F is a polynomial of degree 6. f has a root of multiplicity

2
at
r
=
3
, a root of multiplicity
3
at
r
=
1
,
f
(
−
5
)
=
−
29721.6
, and
f
(
−
10
)
=
0
. Find an algebraic equaton for
f
.

Answer 1

f(x) = 0.43 *
`(x - 3)^(2)` *
`(x-1)^(3)`*(x + 10)

Explanation:

We have a 6th degree polynomial f(x)

r = 3 is a root of f with multiplicity 2

r = 1 is a root of f with multiplicity 3

f(-5) = -29721.6

f(-10) = 0

Then: f(x) = a*((x -3)^2 ) * ((x - 1)^3)*(x + 10)

f(-5) = a * (-8)^2 * (-6)^3 * (5) = -29,721.6

a* (64) * (-216)* 5 = -29,721.6

-a*69,120 = -29,721.6

a = -29,721.6/-69,120

a = 0.43

so

f(x) = 0.43 *
`(x - 3)^(2)` *
`(x-1)^(3)`*(x + 10)