1. roots of polynomials

which of the following polynomials has roots 0, 1, and 2?

help would be wonderful. thanks in advance.

2. Originally Posted by kalter
which of the following polynomials has roots 0, 1, and 2?

help would be wonderful. thanks in advance.
Consider what it means for a number to be a root of a polynomial...

The following graphs depicts a polynomial with roots, x=-2 and x=1.

Consider what happens to the value of y on the graph when x=-2 and when x=1

Also consider that a polynomial of degree n, has at most n real roots.

3. I'm not following. I haven't done this in ages.

4. Originally Posted by kalter
I'm not following. I haven't done this in ages.

Notice on the graph where x=-2 and x=-1, y=0. For a number to be a root of a polynomial, it means that this is where it crosses the x-axis. This means that at that point, y=0.

Does that help?

Also for a polynomial of degree two (this means that the highest exponent is 2)there can be at most 2 real roots.

Same for a polynomial of degree three, there can be at most three real roots.

So considering that you need a polynomial that has those three roots, that should eliminate two of the choices you had.

Does that make sense?

5. Originally Posted by kalter
which of the following polynomials has roots 0, 1, and 2?

help would be wonderful. thanks in advance.
Sub into the equation 0, 1 and 2. If you get 0, it means its a root!

Example
p(x) = $\displaystyle x^2 + x + 2$
Sub in x = 1
p(1) = 1 + 1 + 2 = 4
So 1 is no a root of this equation because p(1) = 4

Equation 2
p(x) = $\displaystyle x^3 - 3x^2 + 2x$
Sub in x = 2
p(2) = $\displaystyle (2)^3 -3(2)^2 + 2(2)$
p(2) = 8 - 12 + 4 which = 0
therefore 2 is a root of $\displaystyle p(x) = x^3 - 3x^2 + 2x$

Done by method of the 'factorial' theorem .