There is some more example of Factorizationusing Identities:- Part II. Let us see some example before moving further:-
Examp:-(e.) 75a3
b2 – 108ab4
Solution: The given expression is not in the form of any identity.
However, its two terms have common factors. Let us rewrite the expression, we
have
(3ab2 × 25a2) – (3ab2 × 36b2) = 3ab2 (25a2 – 36b2)
= 3ab2
{(5a)2 – (6b)2}
=
3ab2 (5a + 6b)(5a – 6b)
By using identity, a2 – b2 = (a +b)(a – b)
Examp:-(f.) 9 – x6 + 2x3y3 – y6 = 9 – (x6 – 2x3y3 + y6)
Solution:-
= 9 - {(x3)2 – 2(x)3(y)3 + (y3)2
= (3)2 – (x3 – y3)2
[On substituting x3 = a, y3 = b in the identity a2 -2ab + b2 = (a – b)2, we have
(x3)2 – 2(x3)(y3) + (y3)2
= (x3 – y3)2]
= {3 + (x3 – y3)}{3
– (x3 – y3)}
Using, a2 – b2 = (a + b)(a – b)
=(3 + x3 –y3)(3 – x3 + y3)
Factorization
by using the identity (x + a) (x +b) = a2 + (a + b)x + ab: – ×
An expression of the type x2 +mx +n is not a perfect square, therefore, cannot be factorized by identities (i) and (ii) used in the preceding topic. This expression also does not fit the identity type (a 2 – b2) either.
However, we can use identity (iv) (x + a) (x +b) = a2 + (a + b)x +
ab.
To factorize this type of expression, we need to find two factor a and
b of the constant term n such that ab = n and a + b = m (the coefficient of the
middle term).
Then, the given expression is factorized in the following way:-
x2 +mx + n = x2 + (a + b) x + ab
= x2
+ ax + bx + ab = x (x + a) + b(x + a)
= (x + a)(x +
b)
This is required factorization.
Exmp:- a. x2 + 14x + 45
Solution:
The constant term 45 can be factorized as 45 = 15 × 3 or 9 × 5,
We’ll consider only those factors of 45, whose sum is equal to 14 (the
coefficient of middle term).
Hence, factor 15 + 3 is not required as 15 + 3 ≠ 14. We have to take
other pair of factors;
i.e 9 +5 = 14
Therefore, the required factor of 45 is 9 and 5.
Putting these factors in the given expression we have,
x2 + 14x + 45 = x2 + (9 + 5)x + (9)(5)
= x2
+ 9x + 5x + (9)(5)
= x2
+ 14x +45
Exmp (b):- x2 – 22x -48
Sol:- The constant term 48 can be factorized as
48 = 8 × 6 or 12 × 4 or 16 × 3 or 24 × 2
Since, 2 – 24 = -22, 2 and – 24 are the required factors of – 48.
So, x2 – 22x – 48 =
x2 + (2 – 24)x + (2) (– 24)
= (x + 2)(x -24)
By using identities, (x +a )(x
+ b) = x2 + (a + b)x + ab
Exmp ©:- x2 -13 x + 36
The constant term 36 can be factorized as
36 = 6 × 6 or 9 × 4 or 12 × 3 or 18 × 2
Since, (– 9) + (– 4) = 1 – 13, - 9
and – 4 are required factors of 36.
Hence, x2 – 13x + 36 = x2 + (- 4 – 9)x +
(-9)(-4)
=x2 – 4x – 9x + (- 9) (-4)
= x(x – 4) – 9(x – 4)
= (x – 4)(x – 9), required factorization.
Negative or Positive Factors of the Constant Term:-
In the given expression x2 +mx + n, where ab = n and a + b
= m, we should follow some rules
regarding positive or negative factors of the constant terms. The rule is following: -
1. If m is positive and n is also positive, then both the factors a
and b of n will be positive.
2. If m is positive and n is negative, then both the factors a and b
of n will have opposite signs. It means one factor will be positive and the
other will be negative.
3. If m is negative and n is positive, then the factors a and b of n
will be negative sign.
4. If m and n are both negative, then the factors a and b of n will
have opposite sings.
Continue..........
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