Factorise Three or more variable Equations
This method is based on the sutra " Lopana Sthapanabyam " which means ' by alternate elemination and retention ' .
Explaining the method with the help of an example .
Example 1 : - Factorise the following : -
x2 + 2y2 + 3.x.y + 2.x.z + 3.y.z + z2.
Answer : -
Step 1 : -
First eleminate terms containing z from the given equation and factorise the remaining equation .
i.e.
put z = 0
x2 + 2y2 + 3.x.y = x2 + 2.x.y + x.y + 2y2
= x.(x + 2.y) + y.(x + 2.y)
= (x + 2.y) . (x + y)
Step 2 : -
Now eleminate the terms containing y and factorise the remaining equation .
i.e.
put y = 0
x2 + 2.x.z + z2 = x2 + x.z + x.z + z2
= x.(x + z) + z.(x + z)
= (x + z) . (x + z)
Step 3 : -
Fill the gaps in the expressions .
i.e.
Fill z in (x + 2.y) &
Fill z in (x + y ) of step 1
Answer is
( x + 2.y + z) . (x + y + z)
Example 2 : - Factorise
3.x2 + y2 - 4.x.y - y.z - 2.z2 - z.x.
Answer : -
Step 1 : -
Put z = 0
3.x2 + y2 - 4.x.y = 3.x2 - 3.x.y - x.y + y2
= 3.x . (x - y) - y.( x - y)
= (x - y) . (3.x - y)
Step 2 : -
Put y = 0
3.x2 - z.x - 2.z2 = 3.x2 - 3.z.x + 2.z.x - 2.z2
= 3.x(x - z) + 2.z(x - z)
= (x - z) . (3.x + 2.z)
Step 3 : -
Fill the gaps
Answer is
(x - y - z) . (3.x - y + 2.z) .
Example 3 : - Factorise
3.x2 + 4.y2 + 7.x.y - 2.x.z - 3.y.z - z2 + 17x + 21y – z + 20.
Answer : -
Step 1 : -
In this eleminate two variables at a time taking independent term as a separate variable .
i.e.
First eleminate y and z put y = 0 & z = 0
3.x2 + 17.x + 20 = 3.x2 + 12.x + 5.x + 20
= 3.x.(x + 4) + 5.(x + 4)
= (x + 4) . (3.x + 5)
Step 2 : -
Now eleminate x and z put x = 0 & z = 0
4.y2 + 21.y + 20 = 4.y2 + 16.y + 5.y + 20
= 4.y.(y + 4) + 5.(y + 4)
= (y + 4) . (4.y + 5)
Step 3 : -
Now eleminate x & y put x = 0 & y = 0
-z2 - z + 20 = -z2 - 5.z + 4.z + 20
= -z.(z + 5) + 4 (z + 5)
= (z + 5) . (-z + 4)
Step 4 : -
Now fill the gaps of various brackets
(x + y - z + 4) . (3.x + 4.y + z + 5)
which is the answer .
Note : - If you want to find the remainder then you have to compare the terms containing x.y & x.z & y.z of gin\ven expression and obtained quatient .
Explaining the method with the help of an example .
Example 1 : - Factorise the following : -
x2 + 2y2 + 3.x.y + 2.x.z + 3.y.z + z2.
Answer : -
Step 1 : -
First eleminate terms containing z from the given equation and factorise the remaining equation .
i.e.
put z = 0
x2 + 2y2 + 3.x.y = x2 + 2.x.y + x.y + 2y2
= x.(x + 2.y) + y.(x + 2.y)
= (x + 2.y) . (x + y)
Step 2 : -
Now eleminate the terms containing y and factorise the remaining equation .
i.e.
put y = 0
x2 + 2.x.z + z2 = x2 + x.z + x.z + z2
= x.(x + z) + z.(x + z)
= (x + z) . (x + z)
Step 3 : -
Fill the gaps in the expressions .
i.e.
Fill z in (x + 2.y) &
Fill z in (x + y ) of step 1
Answer is
( x + 2.y + z) . (x + y + z)
Example 2 : - Factorise
3.x2 + y2 - 4.x.y - y.z - 2.z2 - z.x.
Answer : -
Step 1 : -
Put z = 0
3.x2 + y2 - 4.x.y = 3.x2 - 3.x.y - x.y + y2
= 3.x . (x - y) - y.( x - y)
= (x - y) . (3.x - y)
Step 2 : -
Put y = 0
3.x2 - z.x - 2.z2 = 3.x2 - 3.z.x + 2.z.x - 2.z2
= 3.x(x - z) + 2.z(x - z)
= (x - z) . (3.x + 2.z)
Step 3 : -
Fill the gaps
Answer is
(x - y - z) . (3.x - y + 2.z) .
Example 3 : - Factorise
3.x2 + 4.y2 + 7.x.y - 2.x.z - 3.y.z - z2 + 17x + 21y – z + 20.
Answer : -
Step 1 : -
In this eleminate two variables at a time taking independent term as a separate variable .
i.e.
First eleminate y and z put y = 0 & z = 0
3.x2 + 17.x + 20 = 3.x2 + 12.x + 5.x + 20
= 3.x.(x + 4) + 5.(x + 4)
= (x + 4) . (3.x + 5)
Step 2 : -
Now eleminate x and z put x = 0 & z = 0
4.y2 + 21.y + 20 = 4.y2 + 16.y + 5.y + 20
= 4.y.(y + 4) + 5.(y + 4)
= (y + 4) . (4.y + 5)
Step 3 : -
Now eleminate x & y put x = 0 & y = 0
-z2 - z + 20 = -z2 - 5.z + 4.z + 20
= -z.(z + 5) + 4 (z + 5)
= (z + 5) . (-z + 4)
Step 4 : -
Now fill the gaps of various brackets
(x + y - z + 4) . (3.x + 4.y + z + 5)
which is the answer .
Note : - If you want to find the remainder then you have to compare the terms containing x.y & x.z & y.z of gin\ven expression and obtained quatient .
0 comments:
Post a Comment