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Factorisation by Splitting the Middle Term

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Factorisation of Trinomials of the Form x + qx + r
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To factorise an algebraic expression of the type x + qx + r, split the middle term q(x) in the form of
(a + b)x in such a way that the sum (a + b) is equal to q and their product a × b is equal to the third term r.
Putting, q = a + b and r = ab, we have
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x + qx + r = x + (a + b) x + ab = x + ax + bx + ab = x (x + a) + b(x + a) = (x + a)(x + b)
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Thus, x + qx + r = (x + a)(x + b). Here a and b are rational numbers and a  b  0.
Example 8: Factorise the following:
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(a) x + 9x + 20 (b) x – 8x – 33 (c) x + 25x – 54 (d) x – 13x + 36
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Solution: (a) x + 9x + 20 = x + 4x + 5x + 20
[Here, r = 20, so we have to find
= x(x + 4) + 5(x + 4) = (x + 4)(x + 5) its two factors, whose sum is 9.
We know that
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Or x + 9x + 20 = x + (4 + 5)x + (4 × 5) 20 = 1 × 20 = 2 × 10 = 4 × 5
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Clearly, 4 + 5 = 9 = q
= (x + 4)(x + 5) \ a = 4 and b = 5.]
(b) x – 8x – 33 = x – 11x + 3x – (3 × 11)
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= x(x – 11) + 3(x – 11) [Here, r = –33 = –3 × 11 = –11 × 3
Clearly, –11 + (3) = –8 = q
= (x + 3)(x – 11) \ a = –11 and b = 3]
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(c) x + 25x – 54 = x + (27 – 2)x + (27 × –2)
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= x + 27x – 2x – 54 [Here, r = –54 = (+2) × (–27) = (–2) × 27
Clearly, 27 – 2 = 25 = q
= x(x + 27) – 2(x + 27) \ a = 27 and b = –2]
= (x – 2)(x + 27)
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(d) x – 13x + 36 = x + (–9 – 4)x + 36
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[Here, r = 36 = 9 × 4 = –9 × –4
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= x – 9x – 4x + 36 Clearly, –9 – 4 = –13 = q
\ a = –9 and b = –4]
= x(x – 9) – 4(x – 9) = (x – 4)(x – 9)
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Factorisation of Trinomials of the Form px + qx + r
Example 9: Factorise the following:
(a) 3m – 12m + 12 (b) 10x + x – 3
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Solution: (a) 3m – 12m + 12
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Comparing 3m – 12m + 12 with px + qx + r, we have p = 3, q = –12 and r = 12.
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So, pr = 12 × 3 = 36
We find two factors of 36, that will add up to –12, the coefficient of the middle term.
36 = 1 × 36 = 2 × 18 = 6 × 6 = 3 × 12 [Here q is negative and p × r is positive]
So, the both factors are of (–ve) sign. Thus, we
take a = –6 and b = –6 Note: 1. When both p and r have same

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\ 3m – 12m + 12 = 3m + (–6 – 6)m + 12 sign → Factors a and b will
have like signs.
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= 3m – 6m – 6m + 12 = 3m(m – 2) – 6(m – 2) 2. When both p and r have

= (3m – 6)(m – 2) opposite signs → Factors a
= 3(m – 2)(m – 2) and b will have opposite signs.
303 Factorisation