1. ## Factorisation Help

1. FACTORISE x2 - bx- a2 -5ab -6b2
2. a/b + b/a =1 , find the value of a3 + b3

2. Hello,

Originally Posted by sabyasachi
1. FACTORISE x2 - bx- a2 -5ab -6b2
For the first one :

$\displaystyle x^2-bx-a^2-5ab-6b^2$

$\displaystyle ={\color{red}x}^2-2\cdot {\color{blue}\frac b2} \cdot {\color{red}x}-a^2-5ab-6b^2$

$\displaystyle ={\color{red}x}^2-2\cdot {\color{blue}\frac b2} \cdot {\color{red}x}+\underbrace{\left({\color{blue} \frac b2}\right)^2-\left(\frac b2\right)^2}_{=0}-a^2-5ab-6b^2$

$\displaystyle =\left({\color{red} x}-{\color{blue} \frac b2}\right)^2-\frac{b^2}{4}-a^2-5ab-6b^2$

$\displaystyle =\left(x-\frac b2\right)^2-\left(a^2+5ab+\frac{25}{4} \cdot b^2\right)$

$\displaystyle =\left(x-\frac b2\right)^2-\left({\color{red}a}^2+2 \cdot {\color{red}a} \cdot {\color{blue}\frac{5 \cdot b}{2}}+\left({\color{blue} \frac 52 \cdot b}\right)^2\right)$

$\displaystyle =\left(x-\frac b2\right)^2-\left({\color{red}a}+{\color{blue} \frac 52 \cdot b}\right)^2$

~~~~~~~~~~~~

Now, you have something like x²-y², which is (x-y)(x+y).

3. Originally Posted by sabyasachi
FACTORISE x2 - bx- a2 -5ab -6b2
There is probably a nice little trick to this. But I don't see it.

So let's do it the hard way.

If this is going to factor it will be of the general form:
$\displaystyle (x + c \cdot a + d \cdot b)(x + e \cdot a + f \cdot b)$

When we expand this we get
$\displaystyle x^2 + ce \cdot a^2 + df \cdot b^2 + (c + e) \cdot ax + (f + d) \cdot bx + (cf + de) \cdot ab$

Comparing this to your expression we get
$\displaystyle ce = -1$
$\displaystyle df = -6$
$\displaystyle c + e = 0$
$\displaystyle f + d = -1$
$\displaystyle cf + de = -5$

The solution to this system is fairly simple to derive and gives c = 1, d = 2, e = -1, and f = -3.

So
$\displaystyle x^2 - bx- a^2 - 5ab - 6b^2 = (x + a + 2b)(x - a - 3b)$

-Dan

4. Originally Posted by sabyasachi
a/b + b/a =1 , find the value of a3 + b3
$\displaystyle \frac{a}{b} + \frac{b}{a} = 1$

Multiply both sides by ab:
$\displaystyle a^2 + b^2 = ab$

$\displaystyle a^2 - ab + b^2 = 0$

Now,
$\displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (a + b)(0) = 0$

-Dan

(I see I was faster than Moo this time. )

5. (I see I was faster than Moo this time. )
Actually, I was thinking on it, but I couldn't remember the development of $\displaystyle a^3+b^3$ ^^

And I prefer my method to yours for the first one .

See you next time for another challenge ! xD

(how are you diapers ? )

6. Originally Posted by Moo
(how are you diapers ? )
You know, out of context that question is just sooooo wrong!

-Dan

7. Originally Posted by topsquark
You know, out of context that question is just sooooo wrong!

-Dan
Why ? We both know the context