which of the following polynomials has roots 0, 1, and 2?
help would be wonderful. thanks in advance.
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?
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 .