Alegbra

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February 19th 2006, 11:14 AM
dgolverk

Alegbra

well, I've got some problems that I can't solve, I'm doing the work all the day but I just can't solve those... I would appreciate your help :) even if you can help just in one...
by the way I need it for wednesday so it's pretty urgent...

formula: ax^2+bx+c(trinom)
1.
6x^3+x^2-x+12x^2+2x-2

2.
30x^3-15x^2-14x^2+7x-4x+2

prove:

(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9

(11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400

prove that:

(n^2+2n)(n^2-1)
------------------
24

and last one:
given:
a+b=5
ab=3
How much is:
a^2+b^2
a^2b+ab^2
a^3b+ab^3

Thanks a lot for everyone who can help, even one of all these...
February 19th 2006, 11:48 AM
earboth

Quote:

Originally Posted by dgolverk

well, I've got some problems that I can't solve, I'm doing the work all the day but I just can't solve those... I would appreciate your help :) even if you can help just in one...
by the way I need it for wednesday so it's pretty urgent...

formula: ax^2+bx+c(trinom)
1.
6x^3+x^2-x+12x^2+2x-2

2.
30x^3-15x^2-14x^2+7x-4x+2

prove:

(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9

(11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400

prove that:

(n^2+2n)(n^2-1)
------------------
24

and last one:
given:
a+b=5
ab=3
How much is:
a^2+b^2
a^2b+ab^2
a^3b+ab^3

Thanks a lot for everyone who can help, even one of all these...

Hello,

I cann't do all your problems 'cause I haven't much time now.
to 1. $6x^3+x^2-x+12x^2+2x-2=$
$x(6x^2+x-1)+2(6x^2+x-1)$
$(x+2)(6x^2+x-1)$ now you can factorize the 2nd factor easily.

to 2. $30x^3-15x^2-14x^2+7x-4x+2=$
$15x^2(2x-1)-7x(2x-1)-2(2x-1)$
$(15x^2-7x-2)(2x-1)$ now you can factorize the 1rst factor easily.

$(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9$
set x=7a-8 and y=11-7a. Then you get:
$x^2+2xy+y^2=9\ \Longleftrightarrow \ (x+y)^2=3^2$ Solve for x and y and then resubstitute to get a and b.

$(11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400$ Same procedure: x= 11a+5 and y=11a-15 and you'll get:
$x^2-2xy+y^2=400\ \Longleftrightarrow \ (x-y)^2=20^2$

Now I've to leave the forum. I hope this was of some help to you.

Greetings

EB
February 19th 2006, 01:14 PM
DenMac21

Quote:

Originally Posted by dgolverk

prove that:

(n^2+2n)(n^2-1)
------------------
24

I assume that you want to prove that is dividable with 24, so:
$\frac{{(n^2 + 2n)(n^2 - 1)}}{{24}} = \frac{{(n - 1)n(n + 1)(n + 2)}}{{24}}$

So, n can be any number $24*x,(x \in Q,x \ne 0)$ and it can be n=2 and n=-3 because 24=1*2*3*4.
February 19th 2006, 01:38 PM
DenMac21

Quote:

given:
a+b=5
ab=3
How much is:
a^2+b^2
a^2b+ab^2
a^3b+ab^3

$\begin{array}{l} (a + b)^2 = 5^2 \\ a^2 + 2ab + b^2 = 25 \\ a^2 + b^2 = 25 - 2ab \\ a^2 + b^2 = 25 - 2ab \\ a^2 + b^2 = 25 - 2*3 \\ a^2 + b^2 = 19 \\ \end{array} $

$\begin{array}{l} a^2 b + ab^2 = ab(a) + ab(b) \\ 3a + 3b = 3(a + b) = 3*5 = 15 \\ \end{array} $

$a^3 b + ab^3 = ab(a^2 ) + ab(b^2 )$
$3a^2 + 3b^2 = 3(a^2 + b^2 ) = 3*19 = 57$
February 19th 2006, 02:29 PM
ThePerfectHacker

Let me just explain DenMac's post that
$\frac{n(n+1)(n-1)(n+2)}{24}$ is an integer.
Because $24=4\times 3\times 2\times 1$ .
That means that one of the integers,
$n,n+1,n-1,n+2$ contains a multiple of 4
because there are 4 of them in increasing order. Similarly one of them is also a multiple of 3 and finally a multiple of 2. Thus,
$n(n+1)(n-1)(n+2)$ is a multiple of 24.
-------------------------
If you want a more advanced explanation (I do not know if you learned this thus I do not want to post this one) is through congruences:
Let $x=n-1$
Then, $x+1=n$ , $x+2=n+1$ , $x+3=n+2$ .
Thus, (under modulo 24)
$x\equiv r_1$
$x+1\equiv r_2$
$x+2\equiv r_3$
$x+3\equiv r_4$
Now non-of these are congruent to each other for then,
$x+i\equiv x+j$
Thus,
$i\equiv j$ and $0\leq i,j\leq 3$
Which is impossible.
Thus, the integers $r_1,r_2,r_3,r_4$ must be the integers $1,2,3,4$ in some order by the pigeonhole principle. Thus, $r_1r_2r_3r_4=24$ .
Thus, upon multiplication of the congruences we have that
$x(x+1)(x+2)(x+3)\equiv 24\equiv 0$
Q.E.D.
-----------------------
Finally, there is a theorem from number theory that the product of n-consecutives integers is always divisible by $n!$ which is the case here.
February 20th 2006, 04:24 AM
dgolverk

I'm impressed!

You did great job :)
It's all very helped but I not really understood the answers for these two:
$ (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9 $

and
$ (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400 $

I mean how can I find the X and the Y?
Thanks
February 20th 2006, 11:02 AM
ThePerfectHacker

Quote:

Originally Posted by dgolverk

You did great job :)
It's all very helped but I not really understood the answers for these two:
$ (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9 $

and
$ (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400 $

I mean how can I find the X and the Y?
Thanks

Notice the squares formula $a^2+2ab+b^2=(a+b)^2$
Overhere (in first problem) we have that,
$a=(7a-8)$
$b=(11-7a)$
Thus,
$(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2$
Becomes,
$[(7a-8)+(11-7a)]^2=[3]^2=9$
The same idea is used in the second problem.
February 21st 2006, 02:39 AM
dgolverk

Oh, now I see...

Thanks a lot! :)
February 21st 2006, 02:00 PM
ThePerfectHacker

Welcome