Page 131 - Start Up Mathematics_8 (Non CCE)

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Hence, x + [2 + (–5)]x – 10
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= x + 2x – 5x – 10
= x(x + 2) – 5(x + 2)
= (x – 5)(x + 2) {Taking (x + 2) common}
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(c) x – 21x + 110
To factorize this expression, you need to find p and q such that
p + q = –21, pq = 110
Factors of 110 are 2 × 55, (–2) × (–55), 10 × 11, (–10) × (–11), 22 × 5, (–22) × (–5),
110 × 1, (–110) × (–1)

Out of these factors, only (–10) + (–11) = –21
\ p = –10, q = –11
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Hence, x – 21x + 110
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= x + [–10 + (–11)] x + 110
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= x – 10x –11x + 110
= x(x – 10) – 11(x – 10)
= (x – 11)(x – 10) {Taking (x – 10) common from both the terms}

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Factorization of Quadratic Polynomials of the Form ax + bx + c, a π 1
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Step 1: In the quadratic polynomial ax + bx + c, locate a i.e., coefficient of x , b i.e., coefficient of x and
constant term c.
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Step 2: Find the product ac, i.e., product of coefficient of x and the constant term.
Step 3: Split the coefficient of x, i.e., b in two parts, such that
Ist part + IInd part = b and Ist part × IInd part = ac
Step 4: Factorize the algebraic expression in step 3 by grouping the terms.

Example 13: Factorize the following:
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2
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(a) 5x – 32xy + 12y (b) (3x + 2y) – 6(3x + 2y) – 72
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Solution: (a) 5x – 32xy + 12y 2
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Coefficient of x = 5, coefficient of x = –32y, constant term = 12y 2
To factorize the expression, split the middle term –32y in such a way that the two parts
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sum up to –32y and their product = 60y
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60y can be written as
6y × 10y, (–6y) × (–10y), (–2y) × (–30y), 2y × 30y, 4y × 15y, (–4y) × (–15y), 12y × 5y,
(–12y) × (–5y), 60y × y, (–60y) × (–y)
Out of these factors, only (–2y) + (–30y) = –32y
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2
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\ 5x – 32xy + 12y = 5x + (–2y – 30y)x + 12y 2
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= 5x – 2xy – 30xy + 12y 2
= x(5x – 2y) – 6y(5x – 2y)
= (x – 6y)(5x – 2y) {Taking (5x – 2y) common}
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(b) (3x + 2y) – 6(3x + 2y) – 72
Put 3x + 2y = z
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So, the given expression becomes z – 6z – 72
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