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

Jul 20, 2016

$2 {x}^{3} - 6 {x}^{2} - 12 x + 16$.

#### Explanation:

The degree of reqd. 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 $- 2 , 1 \mathmr{and} 4$.

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

Similarly, other zeroes give us factors $\left(x - 1\right) \mathmr{and} \left(x - 4\right)$

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 $k \ne 0$.

In view of above, we can suppose that,

$p \left(x\right) = k \left(x + 2\right) \left(x - 1\right) \left(x - 4\right)$. Expanding the $R . H . S .$, we have,

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

But this will give us the leading co-eff$= k$, which is given to be $2$, so, $k = 2$.

Hence the poly.$p \left(x\right) = 2 \left({x}^{3} - 3 {x}^{2} - 6 x + 8\right) = 2 {x}^{3} - 6 {x}^{2} - 12 x + 16$.

Enjoy Maths.!