Question

How do you find a polynomial of degree three that has zeroes \[ - 3,0,1\]?

Answer 1

Hint: In the given question, we have been given that there is a polynomial of some degree. Then we have been given the zeros of the polynomial. We have to find the polynomial. We are first going to represent the zeroes as independent polynomials by making each of them equal to separate monomials. Then we are going to multiply the calculated monomials, simplify them, and we get our answer.

Complete step by step answer:
We have been given that there is a polynomial of degree \[3\] and it has zeroes \[ - 3,0,1\].
It means that say the polynomial is represented in terms of variable \[x\], then,
\[x = - 3\], \[x = 0\] and \[x = 1\]
Hence, \[\left( {x + 3} \right),x,\left( {x - 1} \right)\] are the factors of the polynomial. And since the polynomial is of degree three, these three are the only factors.
So, \[p\left( x \right) = \left( {x + 3} \right)\left( {x - 1} \right)\left( x \right)\]
Multiplying and solving,
\[p\left( x \right) = {x^3} + 2{x^2} - 3x\]

Note: In the given question, we had to derive the equation of the polynomial whose zeroes were given. We did that by equaling the zeroes to their corresponding monomials and then multiplying them. The point where a lot of students make mistakes is when they convert the zeroes to monomials, they forget to change the sign of the zeroes. Then, the chances of making an error while multiplying the monomials is high too. So, care must be paid to these things.