How do you write a polynomial in standard form given the zeros x=-6, 2, and 5?

2 Answers
Apr 3, 2016


A polynomial with zeros #x=-6,2,5# is #x^3-x^2-32x+60=0# in standard form.


You are given the following information about the polynomial: zeros.
Definition of zeros: If x = zero value, the polynomial becomes zero. i.e. If you plug in -6, 2, or 5 to x, this polynomial you are trying to find becomes zero.

For the polynomial to become zero at let's say x = 1, the polynomial needs to contain the following term:
#(x-1) = 0#
This looks too simple, and polynomials are usually bigger. They are usually quadratic, cubic, quartic, etc.
However, those messy polynomials all have #(x-1)=0# hidden in them if one of their zeros is 1.

Because we have 3 zeros, we write the same kind of equation for each zero.
#(x+6) = 0#
#(x-2) = 0#
#(x-5) = 0#

You multiply all of them to get the following equation.
If you are wondering why we multiply them, I'll give you a hint. Zero multiplied by any number is zero. So if one of the three equations above becomes zero, does it matter what values the other two equations give?

Expand the equation:

And there's your answer!

One extra note: a more complete answer would be #a(x^3-x^2-32x+60=0)#. If this is homework, the #a# is probably not necessary as an answer. I am just telling you that you can multiply the whole polynomial by some number and not affect the zeros.

Apr 3, 2016


The polynomial equation with roots a, b and c is #x^3-(a+b+c)x^2+(bc+ca+ab)x-abc=0#. Answer: #x^3-13x^2+52x-60=0#-


The cubic equation with roots a, b and c is
#(x-a)(x-b)(x-c) =x^3-(a+b+c)x^2+(bc+ca+ab)x-abc=0#

Here (a, b, c) = (6, 2, 5).
a+b+c+ 13, bc+ca+ab=52 abd abc=60