Third-degree, with zeros of -2, -1, and 3, and passes through the point (2,9)

Answer 1

Hi Joelle!

You probably know by now that the factored form of a polynomial shows us where the "zeros", or x-intercepts, are.

Knowing that y=0 for each zero/solution/x-intercept/etc., we can write a basic form for the polynomial:

f(x)= a (x+2)(x+1)(x-3)

1. This is third degree because we'll get a term with x to the third power.
2. I put a in front of the factors so we can manipulate it to make sure the function passes through the point (2.9).

Substitute 2 for x and 9 for f(x) to get 9 = a (2+2)(2+1)(2-3). Simplify the right hand side to get 9 = -12a.

Divide both sides by -12 to get a = -3/4

Now we have f(x)= -3/4 (x+2)(x+1)(x-3).

Multiply the three factors on the right and combine like terms to get f(x)= -3/4 (x³-7x-6).

Multiply each factor by -3/4 to get f(x)= (-3/4x³ (-21/4)x+9/2).

Please let me know if you have any questions about the steps here.

Have fun mathing!

Liz Z.