Method 16

Finding the H. C .F. of two,three or more degree equation

This method is based on the sutra " Lopana Sthapana " process which means ' elemination and retention or alternate destruction of the highest and the lowest powers '

Explaining the method with the help of an example .

Example 1 : - Find the H.C.F. of
x² + 5x + 4 and x² + 7x + 6.

Answer : -

Just subtract one expression from the other and take out the common from the result .
i.e.
x² + 5x + 4
x² + 7x + 6
-2x -2

Take out common from the result
i.e. - 2

Hence the H. C . F . of the two is

(x + 1) .

Example 2 : - Find H.C.F. of 2x² – x – 3 and 2x² + x – 6 .

Answer : -

2x² – x – 3
2x² + x – 6
-2x + 3

Taske -1 out and so

2x - 3 is the H. C. F. of the two expressions .


Example 3 : - Find the H. C. F. of
x³ – 7x – 6 and x³ + 8x² + 17x + 10.

Answer : -

x³ – 7x – 6
x³ + 8x² + 17x + 10
-8x² -24x -16

Take common out from the result .
i.e.
-8

Hence the H. C. F. of the two expressions is

x² + 3x + 2

Algebraic Proof :

Let P and Q be two expressions and H is their H.C.F. Let A and B the Quotients
after their division by H.C.F.

P = A and Q = B
H H

So P = A.H & Q = B.H

(P + Q) = (A.+ B) . H & (P - Q) = (A - B) . H

Hence we see that
whatever is the H. C. F. of P & Q The same is that of (P + Q) .

Exercises : -

Find the H.C. F. of the following :
1.) x² + 2x – 8 & x² – 6x + 8
2.) x³ – 3x² – 4x + 12 & x³ – 7x² + 16x - 12
3,) x³ + 6x² + 11x + 6 & x³ – x² - 10x - 8
4.) 6x⁴ – 11x³ + 16x² – 22x + 8 &
6x⁴ – 11x³ – 8x² + 22x – 8.
5.) x³ + 6x² + 5x – 12 &
x³ + 8x² + 19x + 12.