Page 3 - LN

3
2
Let f(x) = x – 2x – 5x + 6
So, (x - 1) is a factor of f(x)
Dividing f(x) by (x – 1), we get,

Quotient = q(x )
2
= x – x - 6
Thus, f(x) = (x – 1)q(x)
2
= (x – 1)(x – x – 6)
2
Splitting the middle term of x – x – 6, we get,
2
= [x – 3x + 2x – 6]
= [x(x – 3) + 2(x – 3)]
= (x – 3)(x + 2)
So, q(x) is zero at 3 and –2.
So, the other two zeroes of the polynomial are 2 and –3.

Example:

2
4
3
Find all the zeroes of the polynomial f(x) = 3x + 6x – 2x – 10x – 5, if two of its
zeroes are –1 and –1.
Since, –1 and –1 are two zeroes of f(x).
Therefore, (x + 1)(x + 1) is a factor of f(x)
2
x + 2x + 1 is a factor of f(x)
2
On dividing f(x) by (x + 2x + 1), we have,

Therefore, by division algorithm, we have,
2
2
f(x) = (x + 2x + 1)(3x – 5)
2
2
2
= (x + 2x + 1)[( x) – ( ) ]
2
2
2
= (x + 2x + 1)[( x – )( x + )] (Using identify a – b = (a – b)(a + b))
For other zeroes of f(x), we have,