Page 59 - SM inner class 8.cdr

2
2
2
2
)
(
= (x 2 + x y 2 ) (x 2 - x y 2 ) a - b = ( a + b a - b)]
[ since a = x and b = xy;
E xercise 7.2
1. Factor the given expres sions using iden tity:
2
2
2
4
+
(i) m + 8 m 16 (ii) 4x - 4x + 1 (iii) x + y 9 4 + 6 x y 2
2
4
2
(iv) (a - 8 a b + 16 b 4 ) 18 (v) 256 - x 2 - 2xy y 2
-
-
2. Factor ize using the formula of differ ence of two squares.
2
2
2
2
+
-
(i) x - y 2 (ii) a -( b c) 2 (iii) l -( m n) 2 (iv) 49x - 16
4
2
2
2
(v) x - 3 x y + y 2 4 (vi) 4(a b+ ) - 9(a b ) 2
-
4
3. Factor completely using the formula of differ ence of two squares: m - n 4
4. Factorisation by splitting the middle
2
term 4) Group the terms to 6x + 15x + 4x + 10
form pairs - the first two 3x ( 2x + 5) + 2 2x + 5)
(
We factorise the quadratic expression of the terms and the last two
5
2
2
form ax + bx + c by splitting the middle term. terms. Factor each pair = (3x + ) (2x + )
2
6,
consider x + 5 x + by finding common
factors.
Observe that this expression is not of the type :
5) Factor out the shared (3x + 2 ) (2x + ) 5
2
(a + ) b 2 or (a - ) b , i.e., they are not perfect (common) binomial
squares. parenthesis.
2
Here, in x + 5 x + 6, the term 6 is not a perfect Example 13: Find the factors of 6x - 13x + 6
2
square. So, this expressions obviously also do Solution: 6x - 13x + 6
2
2
not fit the type (a - b 2 ) either. They, however, a ´ c = Product of 6 and 6
2
)
seem to be of the type x + ( a + b x + ab. We
= 36
may therefore, try to use Identity 4.
Factors of 36 = (2,18); (3,12); (4,9)
Quadratic Factorization using Splitting of
Only the factors 4 and 9 gives 13 ® (4 + 9)
Middle Term which is x term is the sum of two
For -13 , both the factors should have
factors and product equal to last term.
negative sign.
To Factor the form : Factor : 6x + 19x + 10 So, – 4 – 9 = - 13
2
2
ax + bx + c
2
Now, 6x - 4x - 9x + 6
1) Find the product of 6 ´ 10 = 60
(
(
x
1st and last term (a ´ ) c . Þ 2 3x - 2) - 3 3x - 2)
2) Find the factors of 60 Þ (3x - 2 ) (2x - ) 3 are the factors.
in such way that factors of 60 are (12,5) ;
addition or subtraction (15,4) ; (60,1); (6,10); 2
of that factors is (30,2) [Here we take 15,4 Example 14. Factorise x + 9 x + 18
the middle term (19x ) as the 15 + 4 = 19, which Solution:
(Splitting of middle is our middle term.
term) Step 1. Find the two numbers whose
product is the constant term, i.e.
2
3) Write the center term 6x + ( 15 + 4 x + 10 (18) and sum is the coefficient of x,
)
using the sum of the = 6x 2 + 15x + 4x + 10
two new factors, i.e. ( )a .
including the proper Step 2. Since the product is positive, there
signs. fore both the factors of 18 will be
either positive or negative.
Step 3. Here the sum is positive so both
the factors of 18 will be positive.
59
Mathematics In Focus - 8