PLEASE HELP!!
1. FACTORISE x2 - bx- a2 -5ab -6b2
2. a/b + b/a =1 , find the value of a3 + b3
Hello,
For the first one :
Try to get quadratic forms.
$\displaystyle x^2-bx-a^2-5ab-6b^2$
$\displaystyle ={\color{red}x}^2-2\cdot {\color{blue}\frac b2} \cdot {\color{red}x}-a^2-5ab-6b^2$
$\displaystyle ={\color{red}x}^2-2\cdot {\color{blue}\frac b2} \cdot {\color{red}x}+\underbrace{\left({\color{blue} \frac b2}\right)^2-\left(\frac b2\right)^2}_{=0}-a^2-5ab-6b^2$
$\displaystyle =\left({\color{red} x}-{\color{blue} \frac b2}\right)^2-\frac{b^2}{4}-a^2-5ab-6b^2$
$\displaystyle =\left(x-\frac b2\right)^2-\left(a^2+5ab+\frac{25}{4} \cdot b^2\right)$
$\displaystyle =\left(x-\frac b2\right)^2-\left({\color{red}a}^2+2 \cdot {\color{red}a} \cdot {\color{blue}\frac{5 \cdot b}{2}}+\left({\color{blue} \frac 52 \cdot b}\right)^2\right)$
$\displaystyle =\left(x-\frac b2\right)^2-\left({\color{red}a}+{\color{blue} \frac 52 \cdot b}\right)^2$
~~~~~~~~~~~~
Now, you have something like x²-y², which is (x-y)(x+y).
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