Online Course Discussion Forum

Math Challenge I-C Handout 2 Questions

 
 
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Math Challenge I-C Handout 2 Questions
by Christina Peng - Monday, 25 September 2017, 3:27 PM
 

I have a lot of questions idk how to do XP, the patterns seem confusing and I have no idea what the Factor Fact(Or Theorem) is about... I tried to start the questions and I put my steps underneath the question. (That is if I knew how to start it, and I didn't for most of them) Pls help if you have any idea....


Question 2.14:

Let KK be a real number and NN be a positive even integer. Is it true that x^N=KxN=K has 22 solutions?

Question 2.15:

Let p(x)=x^2+bx+cp(x)=x2+bx+c, so p(x)p(x) is a quadratic. Suppose the two solutions to p(x)=0p(x)=0 are 33 and 4−4. What is the factored form of p(x)p(x)?

Question 2.16:

The equation 3x^3+18x^236x+24=0−3x3+18x2−36x+24=0 has one integer solution. What is this solution?

-3(x^3-6^2+12x-8)=0...

Question 2.20:

Which of the following is NOT a factored form of x^61x6−1?


Question 2.26:

Factor the expression 27x^3+343

27x^3+7^3=(27x+7)(27x^2-27x*7+7^2)= (27x+7)(27x^2-189x+49)= (27x+7)7(3x^2-27x+7)

= (27x+7)7(3x-1)(x... idk


Question 2.29:

Factor the expression by grouping x^3+3x^2+2x+6


Question 2.30:

Solve the equation xy3x+4y12=0xy−3x+4y−12=0 for xx and yy.

(Edited by John Lensmire - original submission Thursday, 21 September 2017, 10:17 PM:  Reason, cleaned up the problems a little)

 
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Re: Math Challenge I-C Handout 2 Questions
by John Lensmire - Monday, 25 September 2017, 3:41 PM
 

Here are some hints!

For 2.14:

Whenever you are confused by a problem, it is good to start looking at a few specific cases.

  • Let's first look at some with $N=2$:  what can we say about the equations $x^2 = 2$, $x^2 = 4$, $x^2 = 0$, or $x^2 = -4$?
  • We could still look at more examples with $N=4:  what can we say about the equations $x^4 = 2$, $x^4 = 4$, $x^4 = 0$, or $x^4 = -4$?

In general, looking at a few different examples, trying out varying range of values. For example, here we are told $K$ is a real number, but it might be positive, negative, rational, irrational, etc.

For 2.15:

The key here is the Factor Theorem/Fact, but don't be scared by this, it's easier than you think! Consider any question we solve, for example $x^2+3x+2 = 0$ factors as $(x+2)(x+1) = 0$ so we know $x = -2$ or $x = -1$. Do you notice a pattern between the factored form:  $(x+2)\times (x+1)$ and our final solutions $-2$ and $-1$?

For 2.16, 2.20, and 2.26:

Keep in mind our patterns here, for cubes $(a+b)^3$, difference of two squares or difference of two cubes, and sum of two cubes.

  • Your first step in 2.16 is good, can you then use one of the factoring formulas/patterns to take the next step? 
  • For 2.20, keep in mind we can write $x^6 - 1 = (x^2)^3-1^3$ but also $x^6 - 1 = (x^3)^2 - 1^2$.
  • For 2.26, you have the right idea, but just be careful how you're using the formula.

For 2.29:

Try to group as $(x^3 + 3x^2) + (2x+6)$. The second group $2x+6$ has a common factor of $2$ so can be written as $2(x+3)$. Can you do something similar for the first group?

For 2.30:

You can try to do a similar thing to Reverse FOIL here, try $xy - 3x+4y+12 = (x + \underline{\ \ })(y + \underline{\ \ })$.

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Re: Math Challenge I-C Handout 2 Questions
by Christina Peng - Wednesday, 27 September 2017, 4:54 PM
 

Thank you! I have two more questions...


The equation 3x3+18x236x+24=0−3x3+18x2−36x+24=0 has one integer solution. What is this solution? I wrote 6.

First dividing both sides of the equation by 

3−3 (notice all the terms share a common factor of 33) we get x36x2+12x8=0x3−6x2+12x−8=0. The left-hand side we recognize as (x2)3(x−2)3 so we have (x2)3=0(x−2)3=0 so x=2x=2 is the only solution.

The correct answer is: 2
I got this problem wrong but I can't understand what the correct answer means as soon is it magicially transformed into (x-2)^3.


Also another question is are we allowed to ask math questions that are not in handouts and taught in class?

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Re: Math Challenge I-C Handout 2 Questions
by John Lensmire - Wednesday, 27 September 2017, 11:41 PM
 

Taking it one step at a time, we first note that $-3x^3+18x^2-36x+24$ has a common factor of $3$. Hence we can divide by $-3$ on both sides to get $x^3 - 6x^2 + 12x - 8 = 0$. (We divide by $-3$ so we have a positive $x^3$.)

At this point it is a little trickier to see the pattern. We notice that $x^3 = (x)^3$ and $-8 = (-2)^3$, so we want to start examining our patterns involving cubes. From here we try $a = x$ and $b=-2$ in $(a+b)^3$. This gives $$x^3 + 3(x^2)(-2) + 3(x)(-2)^2 + (-2)^3 = x^3 - 6x^2 + 12x - 8$$ so we are left with $(x-2)^3 = 0$. This is where we get $x = 2$.

Do you mean the example questions covered in the video? Yes, you can ask those questions here. The forums are meant for any questions relevant to the courses.