## Monday, October 26, 2015

### Heuristic Solution: Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

Given $p(x)=x^5+x^2+1$ have roots $r_1,\,r_2,\,r_3,\,r_4,\,r_5$. Let $q(x)=x^2-2$. Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

## Sunday, October 25, 2015

### Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

Given $p(x)=x^5+x^2+1$ have roots $r_1,\,r_2,\,r_3,\,r_4,\,r_5$. Let $q(x)=x^2-2$. Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

If we attempt at the problem using the Vieta's formulas, then we might have to work a bit harder and longer to determine the value of the required product. But, it is a workable approach and let's see how and what procedures are needed in order to reach to the final result.

From the polynomial of $p(x)=x^5-0x^4+0x^3-(-1)x^2+0x-(-1)$, we have:

$r_1+r_2+r_3+r_4+r_5=0$

$r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5=0$

$r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_3r_4+r_1r_3r_5+r_1r_4r_5+r_2r_3r_4+r_2r_3r_5+r_2r_4r_5+r_3r_4r_5=-1$

$r_1r_2r_3r_4+r_1r_2r_3r_5+r_1r_2r_4r_5+r_1r_3r_4r_5+r_2r_3r_4r_5=0$

$r_1r_2r_3r_4r_5=-1$

Our target expression is $(r_1-2)(r_2-2)(r_3-2)(r_4-2)(r_5-2)$, which upon expanding we have:

$(r_1-2)(r_2-2)(r_3-2)(r_4-2)(r_5-2)$

$=(r_1r_2r_3r_4r_5)^2+2(r_1^2r_2^2r_3^2r_4^2+r_1^2r_2^2r_3^2r_5^2+r_1^2r_2^2r_4^2r_5^2+r_1^2r_3^2r_4^2r_5^2+r_2^2r_3^2r_4^2r_5^2)$

$\,\,\,\,\,\,+4(r_1^2r_2^2r_3^2+r_1^2r_2^2r_4^2+r_1^2r_2^2r_5^2+r_1^2r_3^2r_4^2+r_1^2r_3^2r_5^2+r_1^2r_4^2r_5^2+r_2^2r_3^2r_4^2+r_2^2r_3^2r_5^2+r_2^2r_4^2r_5^2+r_3^2r_4^2r_5^2)$

$\,\,\,\,\,+8(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$\,\,\,\,\,+16(r_1^2+r_2^2+r_3^2+r_4^2+r_5^2)+32$

But we know too:

$(r_1+r_2+r_3+r_4+r_5)^2$

$=r_1^2+r_2^2+r_3^2+r_4^2+r_5^2+2(r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5)$

This implies $0=r_1^2+r_2^2+r_3^2+r_4^2+r_5^2+2(0)$, i.e.

[MATH]\color{yellow}\bbox[5px,purple]{r_1^2+r_2^2+r_3^2+r_4^2+r_5^2=0}[/MATH]

$(r_1^2+r_2^2+r_3^2+r_4^2+r_5^2)^2$

$=r_1^4+r_2^4+r_3^4+r_4^4+r_5^4+2(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$0=r_1^4+r_2^4+r_3^4+r_4^4+r_5^4+2(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

$-4(r_1^4+r_2^4+r_3^4+r_4^4+r_5^4)=8(r_1^2r_2^2+r_1^2r_3^2+r_1^2r_4^2+r_1^2r_5^2+r_2^2r_3^2+r_2^2r_4^2+r_2^2r_5^2+r_3^2r_4^2+r_3^2r_5^2+r_4^2r_5^2)$

Ah, I just don't have the motivation to keep going with this most likely the fruitless approach, as it will be testing my patient to the limit to look for the appropriate expression or values for:

$r_1^2r_2^2r_3^2r_4^2+r_1^2r_2^2r_3^2r_5^2+r_1^2r_2^2r_4^2r_5^2+r_1^2r_3^2r_4^2r_5^2+r_2^2r_3^2r_4^2r_5^2$ and

$r_1^2r_2^2r_3^2+r_1^2r_2^2r_4^2+r_1^2r_2^2r_5^2+r_1^2r_3^2r_4^2+r_1^2r_3^2r_5^2+r_1^2r_4^2r_5^2+r_2^2r_3^2r_4^2+r_2^2r_3^2r_5^2+r_2^2r_4^2r_5^2+r_3^2r_4^2r_5^2$

It's high time to wise up to stop with this unproductive approach and think for alternative.

I will show you the solution provided by my dearest math mentor from the U.K. in my next blog post, in the mean time, I hope you could make the best of this opportunity to attempt at the solution, who knows, perhaps you could come up with the heuristic solution? :D

## Tuesday, October 13, 2015

### Factorize $x^5+x^4+1$

Factorize $x^5+x^4+1$.

If you have never come across this problem before, I'll bet you dollars to donuts you might reply resolutely that $x^5+x^4+1$ could not be factorized. :D

But, fact is, it can. And one doesn't have to have a crystal ball to "see" the factors of $x^5+x^4+1$. We can work it out, slowly but surely!

## Saturday, October 10, 2015

### Learn To Think Mathematically

Train new mode of thinking and hence deduce something useful as problem solving trick.

It's a breeze if you're asked to factor

$a^2-b^2$

since it's obviously a difference of square that factored as $a^2-b^2=(a+b)(a-b)$.

## Friday, October 9, 2015

### How Learning Patterns Leads to Brighter Students?

Do you know we could make optimal use of the things that we have been told they exist since we were primary school students such as the table that contains the very common info, like the table of squares?

$x$         $x^2$

$1$          $1$
$2$          $4$
$3$          $9$
$4$        $16$

## Wednesday, October 7, 2015

### Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

It's highly unlikely that there is a way to factor $65(a^3b^3+a^2+b^2)-81(ab^3+1)=0$ and when we're asked to solve for the positive integer solutions, we would always think of treating the given equation as a quadratic function and then set its discriminant greater than or equal to zero, but this can't be the case here:

## Monday, October 5, 2015

### IMO Algebra Problem: Find maximum of a₁+a₂+a₃+a₄-a₁a₂-a₁a₃-a₁a₄-a₂a₃-a₂a₄-a₃a₄+a₁a₂a₃+a₁a₂a₄+a₁a₃a₄+a₂a₃a₄-a₁a₂a₃a₄

Find the maximum of

$a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

where $|a_i|\le 1,\,i=1,\,2,\,3,\,4$.

## Thursday, October 1, 2015

### Blaise Pascal: Little Of Everything

It's very important to know a little of everything, as it could help us to become one of the most stand out competitors who has the most competitive edge and that we cannot be replaced easily by anyone else.

Sound too good to be true? It turns out it was. :D