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

1 Answer
Jul 20, 2016

2x^3-6x^2-12x+16.

Explanation:

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

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

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

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

In view of above, we can suppose that,

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

p(x)=k(x^3-3x^2-6x+8)

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

Hence the poly.p(x)=2(x^3-3x^2-6x+8)=2x^3-6x^2-12x+16.

Enjoy Maths.!