Page 161 - ICSE Math 8

EXERCISE 14.4

Factorize the following.
2
2
2
(a) 1 + 2x + x (b) 9x – 24xy + 16y 2 (c) (x – 2) – 16
1
2
2
2
2
2
(d) l + l + (e) 9c – a + 4ab – 4b (f) –4x + 4xy – y 2
4
4
4
2
2
(g) a – b 4 (h) x + 4x + 4 (i) (x + 3) – 8(x + 3) + 16
2
2
2
(j) 49 – (x + 5) 2 (k) 9x – 24x + 12 (l) 4x + 12xy + 9y – 8x – 12y
Factorization of Quadratic Polynomial By Splitting the Middle Term
2
Step 1: In the quadratic polynomial x + ax + b, find a (coefficient of x) and b (constant term).
Step 2: Find two numbers p and q, such that p + q = a and pq = b.
2
2
Step 3: Split the middle term as the sum of px and qx, i.e., x + (p + q)x + b fi x + px + qx + b.
Step 4: Factorize the algebraic expression by grouping.
Example 13: Factorize the following.
2
2
(a) x + 5x + 6 (b) x – 3x – 10
2
Solution: (a) x + 5x + 6
Find p and q, such that p + q = 5, pq = 6.
Factors of 6 are 2 × 3, (–2) × (–3), 6 × 1, (–6) × (–1)
Out of these factors, only 2 + 3 = 5, \ p = 2, q = 3
2
2
Hence, x + (2 + 3)x + 6 = x + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 3)(x + 2) {Taking (x + 2) common}
2
(b) x – 3x – 10
Find p and q such that p + q = –3, pq = –10.
Factors of –10 are (–10) × 1, 10 × (–1), 2 × (–5), (–2) × 5.
Out of these factors, only 2 + (–5) = –3.

\ p = 2, q = –5
2
Hence, x + [2 + (–5)]x – 10
2
= x + 2x – 5x – 10 = x(x + 2) – 5(x + 2)
= (x – 5)(x + 2) {Taking (x + 2) common}

EXERCISE 14.5

Factorize the following algebraic expressions by splitting the middle term.
2
2
2
(a) x – 21x + 108 (b) 40 + 3p – p (c) 9x – x – 20 (d) (x + 7)(x – 10) + 16
2
2
2
2
(e) x – x – 56 (f) x – 11x + 24 (g) 9a – 6a + 1 (h) 2a – 17a – 30
2
2
(i) 2x + 7x – 4 (j) 5a + 13a + 6
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