**Module 4 Quiz**

3 party-goers in the corner are having an intense argument. You walk over to settle the debate.

They are discussing a function g(x). These are your notes.

Professor McCoy: She says 2 is a zero of g(x) because long division with (x+2) results in a remainder of 0.

Ms. Guerra: She says 2 is a zero of g(x) because g(2)=0

Mr. Romano: He says 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.

You correct the reasoning of any inaccurate reasoning by the party-goers.

My justification

2. The remainder theorem states that when the opposite of the constant from the binomial divisor is substituted into a function for x, the result is the remainder. This simply means that if given a polynomial function g(x) and a number a, if (x-a) is a factor of g(x), then a is a zero of the polynomial. To prove that the binomial really is a factor of g(x): using long division with (x+2) as the divisor, using synthetic division with -2 as the divisor, or even substituting 2 into g(x) to equal 0 are all plausible scenarios. So in conclusion, all three party-goers were right in theory; however Professor McCoy and Ms. Guerra were right, while Mr. Romano was partially wrong because he would really need to use -2 as the divisor using synthetic division.

3. To graph a fourth-degree polynomial function, you need to first find the x-intercepts; for example you could use the Rational Root Theorem. After finding the x-intercepts, the next best thing to do is to find the y-intercept by plugging 0 into x and solving for y. Once you have both your x-intercepts and y-intercept, graph them. Now, because you are trying to graph a fourth-degree polynomial, both ends of the graph will point in the same direction. Depending on if the function is positive or not will determine which way the ends will point. If it’s positive, the ends will face up: if negative the ends will face down. From there just draw the graph and connect the dots.

Dr. Collier

**The Party**

A mysterious box is delivered to the dinner party you are attending with the following label...

1. The correct factors of f(x) are (x-3)(x+2)(x+4). I figured this when I first tried to factor the function by grouping and finding the GCF. However, that didn’t really work, but I did find out that (x-3) was definitely a factor. So I followed suit and synthetically divided x^3+3x^2-10x-24 by x-3, getting x^2+6x+8. Finally I factored the trinomial and retrieved the answer of (x-3)(x+2)(x+4)

mysterious box

Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen has four real zeros.

She asks you if it is possible to create a fourth-degree polynomial with only two real zeros.

You reply to her saying...

4. It is definitely possible for a fourth-degree polynomial to have only two real zeros. This is possible because that would mean that the fourth-degree polynomial would also have 2 complex zeros. If a function has complex roots, those complex roots will occur as a conjugate pair. A complex root will have an imaginary i in it. If a function has two real zeros, then it will have two complex zeros. However, a function may never have 3 real zeros and 1 complex zero.

Mr. Romano

Professor McCoy

Mrs. Collins

Dr. Collier summons you over to his table. He wants to demonstrate the graph of a fourth-degree polynomial function, but his graphing calculators' batteries have died.

You explain to Dr. Collier how to create a rough sketch of a graph of a fourth-degree polynomial function.

You have been invited to a fancy dinner party along with distinguished guests varying from math professors, engineers, and financial analysts.

Ms. Guerra