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

Answer:

f(x) = 0.43 * $(x - 3)^{2}$  * $(x-1)^{3}$*(x + 10)

Step-by-step 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)