J17

Let, x 2 – 6x + 8 = (x + a)(x + b).
Thus. X 2 – 6x + 8 = x 2 + (a + b) + ab.
Then, (a + b) = -6 and ab = 8
Two numbers which add up to -6 and whose product is 8 are -2 and -4.
So, x 2 – 6x + 8 = x 2 + (-2 + -4)x + 8
= x 2 – 2x – 4x + 8
= x(x – 2) – 4(x – 2)
= (x – 2)(x – 4).
Example
Factorize x 2 – x – 30
Solution
Remember that x 2 – x – 30 = x 2 – 1x – 30
Let x 2 - x – 30 = (x + a)(x + b)
= x 2 + (a + b)x + ab.
Thus, (a + b) = -1 and ab = -30. 5 and -6 satisfy both of these equations.
So, x 2 – x – 30 = (x + 5)(x – 6)
Example
Factorize x 2 + 3x – 28.
Solution
Let x 2 + 3x – 28 = (x + a)(x + b) = x 2 + (a + b)x + ab
Thus, (a + b) = 3 and ab = -28.
The factors of -28 whose sum is 3, are 7 and -4.
So, x 2 + 3x – 28 = (x + 7)(x – 4)
Example
Factorize:
(a) x 2 + 12x + 36 (b) x 2 – 8x + 16
Solutions
(a)
Let x 2 + 12x + 36 = (x + p)(x + q) = x 2 + (p + q)x + pq
Thus, p + q = 12 and pq = 36.
The factors of 36 whose sum is 12, are 6 and 6.
So, x 2 + 12x + 36 = (x + 6)(x + 6) = (x + 6) 2
(b) Let, x 2 – 8x + 16 = (x + p)(x + q) = x 2 + (p + q)x + pq
Thus, p + q = -8 and pq = 16
The factors of 16 whose sum is -8, are -4 and -4.
So, x 2 – 8x + 16 = (x – 4)(x – 4) = (x – 4) 2
Note: x 2 + 16x + 36 and x 2 – 8x + 16 are perfect squares because their factors are
identical.