TRINOMIAL OF THE TYPE ax²+ bx + c
There can be two Cases -
1) The coefficient of x² is unity, that is, the expression is of the form x² + bx + c
2) The coefficient of x² is not unity.
Case.1) To factorize an expression of the type x² + bx + c, we find two factors p and q of c such that –
So, pq = c & p + q = b
Then, x² + bx + c = x² + (p+q)x + pq
= x² + px + qx + pq
= x(x+p) + q(x+p)
= (x+p)(x+q) (Ans.)
Example.1) Factorize a²+ 11a + 30
Ans.) We have to find a pair of factors of 30 = (1, 30), (5, 6), (2, 15)
By inspection we find that, the product and the sum of the factors 6 & 5 are equal to 30 & 11 respectively.
So, a²+ 11a + 30 = a²+ 6a + 5a + 30
= a (a+6) + 5(a+6)
= (a+6) (a+5) (Ans.)
Example.2) Factorize a²+ 12a + 20
Ans.) We have to find a pair of factors of 30 = (1, 30), (5, 6), (2, 15)
By inspection we find that, the product and the sum of the factors 6 & 5 are equal to 30 & 11 respectively.
So, a² + 11a + 30 = a² + 6a + 5a + 30
= a (a+6) + 5(a+6)
= (a+6) (a+5) (Ans.)
Case.2) To factorize an expression of the form ax²+ bx + c, we find two factors p & q of the product ac (the product of the coefficient of x² and the constant term) such that –
pq = ac & p + q = b,
Example.1) Factorize 5a²+ 25a + 30
Ans.) The coefficient of x² = 5 and the constant term = 50
Their product = 5 X 30 = 150
The possible factors of 150 is = (1, 150), (3, 50), (5, 30), (15, 10)
By inspection we find that, the product and the sum of the factors 15 & 10 are equal to 150 & 25 respectively.
So, 5a² + 25a + 30
= 5a² + 15a + 10a + 30
= 5a (a+3) + 10 (a+3)
= (a+3)(5a+10) = 5 (a+3)(a+2) (Ans.)
Example.2) Factorize 7x²+ 27x + 20
Ans.) The coefficient of x² = 7 and the constant term = 20
Their product = 7 X 20 = 140
The possible factors of 140 is = (1, 140), (7, 20), (7, 30), (15, 10)
By inspection we find that, the product and the sum of the factors 15 & 10 are equal to 150 & 25 respectively.
So, 5a² + 25a + 30
= 5a² + 15a + 10a + 30
= 5a (a+3) + 10 (a+3)
= (a+3)(5a+10)
= 5 (a+3)(a+2) (Ans.)