Friday, July 24, 2015

Analysis for Quiz 11: Creative Problem Solving

Analysis for Quiz 11: Creative Problem Solving

Question 1:  Do you think we can factor part of the expression in the given equation $4x^2+5xy-6y^2-4x+3y-1=0$?

A. Yes.
B. No.
C. Perhaps.


We don't have a crystal ball at hand that we could predict the future, and all that we can say to ambiguous problem like this is that its answer would be a "perhaps".

But if you're familiar with the Olympiad math problem of this genre, you would have told me we could almost always factor part of the expression in the given equation. Like $4x^2+5xy-6y^2-4x+3y$ could be factored completely. We will get to factor it out eventually, the whole point of putting this as the first problem in this quiz is to raise your awareness that there's a distinct possibility that part of the expression from the given equation could be factored into something that is useful to us to answer the subsequent problems.

Question 2: Would you group $4x^2-4x$ and $-6y^2+3y$ together if you're ask to factorize $4x^2+5xy-6y^2-4x+3y$?

Yes, the hints are clear that we could factor out $4x$ from $4x^2-4x$ and also $3y$ from $-6y^2+3y$.


It's very tempting for us to select A as the answer for this second question, as the reason is clearly stated in the statement. But, be careful that this could lead to fairly meaningless step if you're not observant enough a problem solver:


But whether this step is useless or useful to us depends on our algebraic manipulating skills, in fact, if you're bright, you could see if we have introduced a $2y$ and another $x$ into the factors $(x-1)$ and $(2y-1)$ respectively, we would end up having one common factor, which is $(2y+x-1)$, so that gives us the idea that we could rewrite $5xy$ so to split it into two in the following fashion:


But, we could also reach to the same conclusion even if we don't factor out $4x$ from $4x^2-4x$ and $3y$ from $-6y^2+3y$ (which the details would be discussed in the subsequent problems).

Therefore both A and B are the correct answers to this second problem, and this has assured us there are many ways to skin a cat.

Question 3:  If you're told to substitute $x=3$ and $y=4$ simultaneously into the given equation $4x^2+5xy-6y^2-4x+3y-1=0$, what would you have observed?

A. We obtained $0-1=0$, which  is ridiculous.
B. $x=3$ and $y=4$ are roots of $4x^2+5xy-6y^2-4x+3y-1=0$.
C. $x=3$ and $x=4$ are roots of $4x^2+5xy-6y^2-4x+3y=0$.


Yes, if we substitute $x=3$ and $y=4$ into the equation $4x^2+5xy-6y^2-4x+3y-1=0$, we would end up getting $0-1=0$, but we should be very flexible and use it for our advantage. If taking the surface value of the new discovery doesn't bring anything useful, then we should adjust of our finding, that must mean something to us!

What if we scrap the constant in the given equation and that would yield a zero after substituting $x=3$ and $y=4$ into it $4x^2+5xy-6y^2-4x+3y=36+60-96-12+12=0$.

That means, $x=3$ and $y=4$ are roots of $4x^2+5xy-6y^2-4x+3y=0$ and C is the correct answer.

Question 4:  If you've picked the choice C for question 2, how would you factorize $4x^2+5xy-6y^2-4x+3y$?

A. Polynomial long division to divide  $4x^2+5xy-6y^2-4x+3y$ by $(x-3)$ .
B. Polynomial long division to divide  $4x^2+5xy-6y^2-4x+3y$ by  $(y-4)$.
C. Polynomial long division to divide  $4x^2+5xy-6y^2-4x+3y$ by  $(x-3)(y-4)$.
D. Neither of the above.


Our common assumption is if $x-a$ is the factor of the polynomial $P(x)$, then we could almost always factor it with the polynomial long division method and it will give us the perfect factored form where $P(x)=(x-a)Q(a)$, where $Q(a)$ is the quotient that we got from the polynomial long division step.

I will show you this is not necessarily be true in all cases.

[MATH]\begin{array}{r}4x+5y\hspace{140px}\\x-3\enclose{longdiv}{4x^2+5xy-6y^2-4x+3y} \\ -\underline{\left(4x^2-12x\right)} \hspace{110px} \\ 5xy-6y^2+8x+3y \hspace{10px} \\ -\underline{(5xy-15y)} \hspace{70px}\\ -6y^2+8x+18y\hspace{20px}\end{array}[/MATH]

[MATH]\begin{array}{r}-6y+5x\hspace{140px}\\y-4\enclose{longdiv}{-6y^2+3y+5xy+4x^2-4x} \\ -\underline{\left(-6y^2+24y\right)} \hspace{110px} \\ -21y+5xy+4x^2-4x \hspace{10px} \\ -\underline{(5xy-20x)} \hspace{40px}\\ 4x^2-4x-21y\hspace{20px}\end{array}[/MATH]

[MATH]\begin{array}{r}\text{I don't think you want to even begin to divide}\\\hspace{20px}\\xy-4x-3y+12\enclose{longdiv}{-6y^2+3y+5xy+4x^2-4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \end{array}[/MATH]

Hence, we could never divide $4x^2+5xy-6y^2-4x+3y$ by any of the suggested divisors above, so D is the answer to this problem.

Question 5:  If you've failed to factorize $4x^2+5xy-6y^2-4x+3y$, what else could you do to factorize it completely by making full use of the info that substituting x=3 and y=4 into 4x²+5xy-6y²-4x+3y yields a zero?

A. I would try to divide $4x^2+5xy-6y^2-4x+3y$ by  $(x+y-7)$.
B. I would try to divide $4x^2+5xy-6y^2-4x+3y$ by  $(xy-12)$.
C. I would try to divide $4x^2+5xy-6y^2-4x+3y$ by  $(4x-3y$).
D. We could not factorize the expression $4x^2+5xy-6y^2-4x+3y$.


Actually we could also determine the correct answer by doing the polynomial long division and with a little work, we can check that C is the answer.

[MATH]\begin{array}{r}x+2y-1\hspace{120px}\\4x-3y\enclose{longdiv}{4x^2+5xy-6y^2-4x+3y} \\ -\underline{\left(4x^2-3xy\right)} \hspace{110px} \\ 8xy-6y^2-4x+3y \hspace{10px} \\ -\underline{(8xy-6y^2)} \hspace{70px}\\ -4x+3y\hspace{10px}\\-\underline{(-4x+3y)} \hspace{5px}\\ 0\hspace{10px}\end{array}[/MATH]

Therefore we get $4x^2+5xy-6y^2-4x+3y=(4x-3y)(x+2y-1)$.

Question 6:  Solve for integer solutions of $4x^2+5xy-6y^2-4x+3y-1=0$.

A. No integer solution.
B. There are $3$ solutions.
C. There are $4$ solutions.
D. There are infinite solutions.


Since $4x^2+5xy-6y^2-4x+3y=(4x-3y)(x+2y-1)$, we then could rewrite the equation $4x^2+5xy-6y^2-4x+3y-1=0$ as:



We're told both $x$ and $y$ are integers, and the only way the product of two integers could yield a $1$ is both must be a $1$.

So we must let $4x-3y=1$ and $x+2y-1=1$. Solving them for both $x$ and $y$ gives $x=\dfrac{8}{11},\,y=\dfrac{7}{11}$, which are not integer solutions.

Therefore, the problem has no integer solution and A is the answer to the problem.

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