Page 130 - Start Up Mathematics_8 (Non CCE)

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(c) x + y – 2(xy + yz – xz) = x + y – 2xy – 2yz + 2xz
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= (x – 2xy + y ) + (2xz – 2yz) (Regrouping)
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= (x – y) + 2z(x – y) {Using p – 2pq + q = (p – q) }
= (x – y + 2z)(x – y)
{Taking (x – y) common from both the terms}
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(d) a – b + a + b = {(a ) – (b ) } + (a + b ) {Using p – q = (p – q)(p + q)}
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= (a + b )(a – b ) + (a + b ) {Taking (a + b ) common from
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= (a – b + 1)(a + b ) both the terms}
EXERCISE 7.4
Factorize the following:
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(a) 1 + 2x + x (b) 9x – 24xy + 16y 2 (c) (x – 2) – 16 (d) l + l + 1
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(e) 9c – a + 4ab – 4b (f) –4x + 4xy – y 2 (g) a – b 4 (h) x + 4x + 4
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(i) (x + 3) – 8(x + 3) + 16 (j) 49 – (x + 5) 2 (k) 9x – 24x + 12 (l) 4x + 12xy + 9y – 8x – 12y
Factorization of Quadratic Polynomial in One Variable
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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.
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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 12: Factorize the following:
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(a) x + 5x + 6 (b) x – 3x – 10 (c) x – 21x + 110
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Solution: (a) x + 5x + 6
To factorize the above expression you need to 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
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Hence, x + (2 + 3)x + 6 = x + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 3)(x + 2) {Taking (x + 2) common}
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(b) x – 3x – 10
To factorize this expression, you need to 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
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