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.

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

$(x-2)*(x-3)*(x-5)=0\\ (x^{2} -3x-2x+6)*(x-5)=0\\ (x^{2} -5x+6)*(x-5)=0\\ x^{3} -5x^{2} -5x^{2} +25x+6x-30=0\\ x^{3}-10x^{2} +31x-30=0$

therefore

the answer Part b) is

the cubic polynomial function is equal to

$x^{3}-10x^{2} +31x-30=0$

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yawa3891 [41]

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