# CLASS-8ALGEBRAIC FACTORIZATION - TRINOMIAL OF THE TYPE - ax²+ bx + c

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.)