Answer:
h(t) = 6t²(t -2)²(t +5) . . . . . factored form
h(t) = 6t⁵ +6t⁴ -96t³ +120t²
Step-by-step explanation:
You want a polynomial h(t) of degree 5 with roots at 0 and 2 of multiplicity 2 and at -5, such that h(1) = 36.
Factors
Each root a t = r with multiplicity m gives rise to a factor ...
[tex](t -r)^m[/tex]
Using the roots and multiplicities give, the factors are ...
[tex]y=(t-0)^2(t-2)^2(t-(-5))=t^2(t-2)^2(t+5)[/tex]
Scaling
The polynomial h(t) will be a multiple of this, scaled by a factor that makes h(1) = 36.
For t=1, the value of y is ...
y = 1²(1 -2)²(1 +5) = 1(1)(6) = 6
To give h(1) the value of 36, we need to multiply y by 36/6 = 6.
h(t) = 6t²(t -2)²(t +5) = 6t⁵ +6t⁴ -96t³ +120t²
__
Additional comment
The total number of roots is equal to the degree, meaning the root at -5 must have multiplicity 1.
