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.