neelam wrote:
hi .. could u pls tell me why does not 1 work here..if you substitute x=1, even then the number is divisible by x+1.
pls explain..
If you substitute x=1,
=> x=1
=> x-1=0
we are not looking to see if x-1 is a factor. We are looking to see if x+1 is a factor. As a simple rule, you can equate the factor (x+1) to zero (x+1=0) and find the value of x (x+1=0 => x=-1) and then subsitute this in the original expression.
lastochka wrote:
apparently my algebraic skills aren't where they should be... I couldn't follow this explanation
I will try to explain as much as I can - from little notes and distant memory.
We generally note algeraic expressions as functions, f(x) -- this means that f(x) is a polynomial in x and for each value of x, f(x) gives you a result.
Rules:
For any polynomial f(x):
1. To find factors, solve: f(x)=0
2. For any value a, f(a) gives the reminder when f(x) is divided by x-a.
#2 inutrn means,
For any factor x-a, f(a) gives 0
For any nonfactor x-a, f(a) gives reminder
Lets take an example: f(x) = x^2-5x+6
To find factors:
equate f(x) = 0,
x^2-5x+6=0
(x-3)(x-2)=0
x=3 or 2
To see if x-3 is a factor:
x-3=0 => x=3
f(x) = x^2-5x+6
f(3) = 3^2-5.3+6
f(3) = 0
So, x-2 is a factor of x^2-5x+6
To see if x+5 is a factor:
x+5=0 => x=-5
f(x) = x^2-5x+6
f(5) = 5^2+5.5+6
f(3) = 56
So, x+5 is NOT a factor of x^2-5x+6.
And when you divide x^2-5x+6 by x+5, you will get a reminder of 56.
I think this should be enough for this topic, but algebra in itself is a ocean.