we know that
For a polynomial, if
x=a is a zero of the function, then
(x−a) is a factor of the function. The term multiplicity, refers to the number of times that its associated factor appears in the polynomial.
So
In this problem
If the cubic polynomial function has zeroes at 2, 3, and 5
then
the factors are
\((x-2)\\ (x-3)\\ (x-5)\)
Part a) Can any of the roots have multiplicity?
The answer is No
If a cubic polynomial function has three different zeroes
then
the multiplicity of each factor is one
For instance, the cubic polynomial function has the zeroes
\(x=2\\ x=3\\ x=5\)
each occurring once.
Part b) How can you find a function that has these roots?
To find the cubic polynomial function multiply the factors and equate to zero
so
\((x-2)*(x-3)*(x-5)=0\\ (x^{2} -3x-2x+6)*(x-5)=0\\ (x^{2} -5x+6)*(x-5)=0\\ x^{3} -5x^{2} -5x^{2} +25x+6x-30=0\\ x^{3}-10x^{2} +31x-30=0\)
therefore
the answer Part b) is
the cubic polynomial function is equal to
\(x^{3}-10x^{2} +31x-30=0\)
