Question

What is a cubic polynomial function in standard form with zeroes 1, –2, and 2?

Answer 1

Starting off with the polynomial in standard form would be extremely difficult, but we can construct one fairly easily with the zeroes we've been given.

We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:

$(x-1)(x+2)(x-2)=0$

Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:

$(x-1)(x+2)(x-2)=0\\ \big[(x-1)x+(x-1)2\big](x-2)=0\\ (x^2-x+2x-2)(x-2)=0\\ (x^2+x-2)(x-2)=0\\ (x^2+x-2)x-(x^2+x-2)2=0\\ x^3+x^2-2x-(2x^2+2x-4)=0\\ x^3+x^2-2x-2x^2-2x+4=0\\ x^3+(x^2-2x^2)+(-2x-2x)+4=0\\ x^3-x^2-4x+4=0$

So, in standard form, our cubic polynomial would be $x^3-x^2-4x+4$