Anybody here taking Algebra 2 for virtual school and has done lesson the Module 4 Quiz?

Can anyone help me with this problem?

4.Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen have four real zeros. She asks you if it is possible to create a fourth-degree polynomial with only two real zeros. Demonstrate how to do this and explain your steps.

Question

Anybody here taking Algebra 2 for virtual school and has done lesson the Module 4 Quiz?

Can anyone help me with this problem?

 4.Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen have four real zeros. She asks you if it is possible to create a fourth-degree polynomial with only two real zeros. Demonstrate how to do this and explain your steps.

mathmale · Answer

You could do this by inventing zeros (2 real and 2 complex), writing the corresponding factors, and then multiplying out the four factors.
To help you get started:  Imagine that your 2 complex zeros are {2-i,2+i,3, 4}.
The corresponding factors are (x-2+i),(x-2-i),(x-3), and (x-4).  Try multiplying these out to obtain a fourth order polynomial with only two real roots.

anonymous · Answer

Like multiplying those four together?

anonymous · Answer

@mathmale

mathmale · Answer

Exactly.  It's easier to multiply (x-3)(x-4) than to multiply (x-2+i)(x-2-I) (although you still do have to do all of this multiplication).  Try multiplying (x-3)(x-4) now.  Use FOIL.

anonymous · Answer

I got x^2-7x+12. @mathmale

anonymous · Answer

Is that correct? @mathmale

mathmale · Answer

Great.  Now let's begin multiplying out the complex factors.
I'd arbitrarily subdivide the info within parentheses as follows:  
(x-[2+i])(x-[2-i]).

mathmale · Answer

Hint:  use this principle:  (a-b)(a+b)=a^2 - b^2.  This is a "special product."

mathmale · Answer

Thus, (x-[2+i])(x-[2-i]) would be written as x^2 - [2+i][2-i].

Have you done this kind of problem before?

anonymous · Answer

I do not believe so.

anonymous · Answer

Well at least not including the brackets in a problem.

mathmale · Answer

Has your teacher (or have your instructional materials) ever discussed "complex numbers"?

anonymous · Answer

Oh yes she has!

mathmale · Answer

Good.  Let's start with a simpler example:

(a+i)(a-i) = a^2 + (what?)

anonymous · Answer

Would I foil in this case?

mathmale · Answer

Recall that (a-b)(a+b)=a^-b^2 when both a and b are real.
But here we have (a+i)(a-i), where a is real and i is the imaginary operator.

Yes, you would use FOIL to multiply those two binomials together.

anonymous · Answer

I believe it is a^2+ai+ai-1 I am pretty sure I am incorrect but I am not sure.

anonymous · Answer

@mathmale

anonymous · Answer

Hello?@mathmale