Question

How do you find the polynomial function with a leading coefficient $2$ that has the given degree and zeroes: degree $3$, zeroes $-2,1,4$?

Answer 1

Find the required polynomial:

Given: Leading coefficient $=2$, degree$=3$ and zeroes $=-2,1,4$

The degree of required polynomial say $p\left(x\right)$ is $3$, and hence by the Fundamental Principle of Algebra, it must have $3$ zeroes. These are given to be $-2,1,4$.

As $-2$is a zero of $p\left(x\right),x-(-2)=x+2$ must be a factor of $p\left(x\right)$.

Similarly, other zeroes give us factors $(x-1)$and $(x-4)$

Degree of $p\left(x\right)$ is $3$, so,$p\left(x\right)$ can not have any other factor except those described above. Of course, $p\left(x\right)$can have a numerical factor, like $k\ne 0$.

So, we can suppose that, $p\left(x\right)=k(x+2)(x-1)(x-4).$

Expanding the R.H.S., we have, $p\left(x\right)=k(x^3-3x^2-6x+8)$, here leading coefficient$=k$

Since the leading coefficient is $k=2$.

$\therefore p\left(x\right)=2(x^3-3x^2-6x+8)$

Hence, the required polynomial function is $p\left(x\right)=2(x^3-3x^2-6x+8)$.