Help with a Pre-Calc Question, hope I'm on the right thread

P

panamarojo1989

New member
Joined
Nov 18, 2018
Messages
9
Hi, I'm stuck on this question, could anyone work it out for me?
 

Attachments

  • IMG_3840.jpeg
    IMG_3840.jpeg
    1.1 MB · Views: 15
You misunderstand what we do. We do not give answers (except in rare cases). We help students solve a problem on his or her own. One way we do that is to give hints.

Hint 1: If a polynomial has only real coefficients and is of degree 2n - 1, it can be factored into n quadratics and one linear factor, all with real coefficients.

Hint 2: If a quadratic has real coefficients, it can be factored into two linear terms with real coefficients or with a pair of conjugate complex numbers.
 
Isn't it clear that x=1 is zero? If 6i is a solution, then what is another solution? What factors do you have from these zeros? These factors will get you down to a quadratic factor.

Show us your work, please!
 
Well I got up to here LOL. You might not believe me but I used to be good at this
 

Attachments

  • Screen Shot 2020-07-05 at 4.33.27 PM.png
    Screen Shot 2020-07-05 at 4.33.27 PM.png
    1,020 KB · Views: 6
1593981951142.png

First, go back and read the relevant section of your textbook. It should tell you that if 6i is a zero, then so is -6i. It should also tell you how to use synthetic division, or long division, to start the process of factoring. Give that a try, and show us how far you can get doing that.

(The work you show does not go in the direction of factoring at all.)
 
Synthetic division! that's right! I don't have a book I'm doing this purely out of fun between my buddies. Ok let me give it a try I used to be good at synthetic division.
 
You usually only want to factor an equation when I side is 0.
So know you have 3 zeros; x=1, x=6i and x=-6i. Then 3 factors must be x-1, x-6i and x+6i. Multiply these out and then divide that factor into the original polynomial using long division. Then factor that quotient. Now you will have the 6 linear factors.
 
Jomo said:
You usually only want to factor an equation when I side is 0.
So know you have 3 zeros; x=1, x=6i and x=-6i. Then 3 factors must be x-1, x-6i and x+6i. Multiply these out and then divide that factor into the original polynomial using long division. Then factor that quotient. Now you will have the 6 linear factors.
Click to expand...
There are, of course, many ways to proceed. I wouldn't have said that x=1 is an obvious zero (though with experience one might know to check for that possibility). You can divide separately by x-6i, then x+6i, using synthetic division; and then see if another zero becomes more obvious; or you can take various shortcuts, some using long division.

panamarojo1989 said:
Synthetic division! that's right! I don't have a book I'm doing this purely out of fun between my buddies. Ok let me give it a try I used to be good at synthetic division.
Click to expand...
That sure looks to me like a course someone is taking! It even has a course number in the current semester. And such software commonly includes access to some sort of teaching ...
 
JeffM said:
Because the student will probably not be able to do so on the test.
Click to expand...
That is exactly my point: why would that be on a test?
If you were testing would you require students to find \(\sqrt{13}\) by hand?
I really doubt that in twenty years that any calculus textbook will have a chapter on techniques of integration.
Small tablet computers with computer algebra systems are going to change mathematics the way calculators did twenty years ago.
Today it is unthinkable to test a student requiring that \(\left(1+\sqrt{33}\right)^{10}\) be evaluated without a calculator.
 
pka said:
That is exactly my point: why would that be on a test?
If you were testing would you require students to find \(\sqrt{13}\) by hand?
I really doubt that in twenty years that any calculus textbook will have a chapter on techniques of integration.
Small tablet computers with computer algebra systems are going to change mathematics the way calculators did twenty years ago.
Today it is unthinkable to test a student requiring that \(\left(1+\sqrt{33}\right)^{10}\) be evaluated without a calculator.
Click to expand...
Oh, I agree with you that it makes absolutely no sense in the modern world to put such things on a test. That does not at all entail that it will not be on a test. "Should not" and "will not" have very distinct meanings. if I had my way, no one would bother any longer with teaching mathematical mechanics. Everything would be word problems; how do you translate this problem into symbols that a computer can cope with? In the meantime, diligent students want and need to pass inane tests. You are talking to future teachers. I am talking to current students.
 
I am concerned about one thing which I mentioned before. Students need to learn how to think! Now if 'the new math' can teach students how to think mathematically then I fully agree with the attitude shown by pka. That would be great that students did not need to learn techniques which a computer/calculator can do as I agree that it is useless to learn solely to be able to get the answer. Using WO can do that for you.
 
The piece of info you're missing is this: If 6i is a zero then (x^2 + 36 ) is a factor. If you can divide, the other part is just a cubic.
 
Last edited:
You must log in or register to reply here.
Top