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
Ratnaker Mehta
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.!

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