Question

What is the standard form of a cubic polynomial function with zeros 3, -6 and 0?

Answer 1

Hint: We will utilise the factored form of the zeros to get the standard form of the cubic polynomial with zeros 3, -4, and 5. First, the function's zero is 3, -4, and 5. All of these values fulfil the function, when we multiply these factor forms, we will get the function.

Complete step by step solution:
We have given roots of the cubic polynomial as: 3, -6 and 0
We will find their corresponding linear factors.
Factors are: $x$ , $(x - 3)$ and $(x + 6)$
We know that the product of all factor is equals to the function, we let the function as y
$ \Rightarrow y = x(x - 3)(x + 6)$
We will now multiply them
$ \Rightarrow y = x \times (x - 3) \times (x + 6)$
We will first multiply the last two factors
$ \Rightarrow y = x \times (x \times (x - 3) + 6 \times (x - 3))$
$ \Rightarrow y = x \times ({x^2} - 3x + 6x - 18)$
We will then multiply the last term by x
$ \Rightarrow y = x \times ({x^2} + 3x - 18)$
$ \therefore y = {x^3} + 3{x^2} - 18x$
Hence, the cubic polynomial having zeros 3, -6 and 0 is ${x^3} + 3{x^2} - 18x$

Note:
Three zeros are always present in a cubic polynomial, and they might be the same or different. Depending on the function, the zeros can be both real and complex. The zeros, when substituted in the provided function, fulfilled it and resulted in a zero response. There are several methods for finding the zeros of a cubic polynomial, but the most common one involves three steps: first, we use the hit-and-try method to obtain one zero; second, we divide our function and solve the quotient obtained using the quadratic formula to obtain another two zeros; and finally, we divide our function and solve the quotient obtained using the quadratic formula to obtain another two zeros.