- Find the HCF of each of the following pairs of polynomials
(i) x2 + 2x – 15 and x2 – 7x + 12
Solution:
x2 + 2x – 15 & x2 – 7x + 12
= x2 + 5x – 3x – 15 = x2 – 4x – 3x – 12
= x (x + 5) – 3 (x + 5) = x (x – 4) – 3 (x – 4)
=(x + 5) (x – 3) = (x – 4) (x – 3)
HCF = (x – 3)
(ii) m2 – 3m + 2 and m2 + m – 6
Solution:
m2 – 3m + 2
= m2 – 2m – m + 2
= m (m – 2) – 1 (m – 2)
= (m – 2) (m – 1)
m2 + m – 6
= m2 + 3m – 2m – 6
= m (m + 3) – 2 (m + 3)
= (m + 3) (m – 2)
HCF = (m – 2)
(iii) x2 – xy – 2y2 and x2 + 3xy + 2y2
x2 – xy – 2y2 = x2 – 2xy + xy – 2y2
= x (x – 2y) + y (x – 2y)
= (x – 2y) (x + y)
x2 + 3xy + 2y2
= x2 + 3xy + xy + 2y2
= x (x + 2y) + y (x + 2y)
= (x + 2y) (x + y)
HCF = (x + y)
(iv) x2 – 2x + 1 and x2 – 3x + 2
x2 – 2x + 1 = (x – 1)2
x2 – 3x + 2 = x2 – 2x – x + 2
= x (x – 2) – 1 (x – 2)
= (x – 2) (x – 1)
HCF = (X – 1)
- Find the HCF of p(x) = (x3 -27) (x2 – 3x +2) and
q(x) =(x2 + 3x + 9) (x2 – 5x +6)
Solution:
p(x) = (x3 -27) (x2 – 3x +2)
(x3 -27) = x3 – 93 = (x – 3) (x2 + 3x + 32)
= (x – 3) (x2 + 3x + 9)
(x2 – 3x +2) = x2 – 2x – x – 2
= x (x – 2) – 1 (x – 2)
= (x – 2) (x – 1)
p(x) = (x – 3) (x2 + 3x + 9) (x – 2) (x – 1)
q(x) = (x2 + 3x + 9) (x2 – 5x +6)
(x2 – 5x +6) = x – 3x – 2 x + 6
=x (x – 3) – 2 (x – 3)
= (x – 3) (x – 2)
q(x) = (x – 3) (x – 2) (x2 + 3x + 9)
HCF = (x – 3) (x – 2) (x2 + 3x + 9)
- Find the HCF of f(x) = x3 + x2 – x – 1 and g(x) = x3 + x2 + x + 1
Solution:
f(x) = x3 + x2 – x – 1
= x3 (x + 1) – 1 (x + 1)
= (x2 – 1) (x + 1)
= (x + 1) (x – 1) (x + 1)
= (x + 1)2 (x – 1)
g(x) = x3 + x2 + x + 1
= x2 (x + 1) + 1 (x + 1)
= (x2 – 1) (x + 1)
HCF = (x + 1)
-
- Find the HCF of p(x) = x4 – 2x3 – 15x2and q(x) = x3 – 9x
p(x) = x4 – 2x3 – 15x2
= x2 (x2 – 2x – 15)
= x2(x2 -5x + 3x – 15)
= x2 [x (x – 5) + 3(x – 5)]
= x2 (x + 3) (x – 5)
q(x) = x3 – 9x
= x (x2 – 9)
= x (x2 – 32)
= x (x – 3) (x + 3)
HCF of p(x) and q(x) = x (x + 3)
- Find the HCF of each of the following triples:
(i) a4 b – ab4 , a4b2 – a2b4 and a2 b2 (a4 – b4)
a4b – ab4 = ab (a3 – b3)
= ab (a – b) (a2 + 2ab +b2)
a4b2 – a2b4 = a2 b2 (a2 – b2)
= a2 b2 (a – b) (a + b)
a2 b2 (a4 – b4) = a2 b2 (a2 + b2) (a2 – b2)
= a2 b2 (a2 + b2) (a – b) (a + b)
HCF = ab (a – b)
(ii) 6(x2 + 10x + 24), 4(x2 – x – 20) and 8(x2 + 3x – 4)
Solution:
6(x2 + 10x + 24) = 6(x2 + 6x + 4 x + 247)
= 6 [x (x + 6) + 4 (x + 6)]
= 2 x 3 (x + 6) (x + 4)
4(x2 – x – 20) = 4 (x – 5x + 4x – 20)
= 4 [x (x – 5) + 4 (x – 5)]
= 22 (x – 5) (x + 4)
8(x2 + 3x – 4) = 23 (x2 + 4x – x – 4)
= 23 [x (x + 4) – 1 (x + 4)
= 23 (x + 4) (x – 1)
HCF = 2 (x + 4)
(iii) a2 (2x2 + 5x + 2), a2b (3x2 + 8x +4)and ab2 (2x2 + 3x – 2)
Solution:
a2 (2x2 + 5x + 2) = a2 (2x2 + 4x + x + 2)
= a2 [2x (x + 2) + 1 (x + 2)]
= a2 (2x + 1) (x + 2)
a2b (3x2 + 8x +4) = a2b (3x2 + 6x + 2x +4)
= a2b [3x(x + 2) + 2 (x + 2)]
= a2b (x + 2) (3x + 2)
ab2 (2x2 + 3x – 2) = ab2 (2x2 + 4x – x – 2)
= ab2 [2x (x + 2) – 1 (x + 2)]
= ab2 (2x – 1) (x + 2)
HCF is a (x + 2) HCF = a (x + 2)
- If the HCF of the polynomials p(x) = (x + 1) (x2 + ax + 4) and
q(x) = (x + 4) (x² + bx +2) is h(x) = x2 + 5x + 4, find the values of a and b.
Solution:
p(x) = (x + 1) (x2 + ax + 4)
q(x) = (x + 4) (x2 + bx +2)
h(x) = x2 + 5x + 4
= x2 + 4x + x + 4
= x (x + 4) + 1(x + 4)
= (x + 4) (x +1)
Hence (x + 1) divides x2 + bx + 2
x2 + bx + 2 = (x + 1) f(x)
Let x = –1
(–1)2 + b (–1) + 2 = 0
–b + 3 = 0
b = 3
Also (x + 4) divides x2 + ax + 4
x2 + ax + 4 = (x + 4) g(x)
Let x = – 4
(–4)2 + a (–4) + 4 = 0
16 – 4a + 4 = 0
4a = 20
a = 5
- If the HCF of the polynomials p(x) = (x – 3) (2x2 – ax + 2) and
q(x) = (x + 4) (x2 – bx – 6) is h(x) = x2 – 5x + 6, find the values of a and b.
Solution:
h(x) = x2 – 5x + 6
= x2 – 3x – 2x + 6
= x (x – 3) – 2 (x – 3)
= (x – 3) (x – 2)
p(x) = (x – 3) (2x2 – ax + 2)
Hence (x – 2) divides 2x2 – ax + 2
2x2 – ax + 2 = (x – 2) f(x)
Let x = 2
2(2)2 – a (2) + 2 = 0
8 – 2a + 2 = 0
2a = 10
a = 5
q(x) = (x + 4) (x2 – bx – 6)
Hence (x – 3) divides (x2 – bx – 6)
(x2 – bx – 6)= (x – 3) g(x)
Let x = 3
33 – b (3) – 6 = 0
9 – 3b – 6 = 0
3b = 3
b = 1
- For what values of a and b the polynomials p(x) = (x2 +5x+6) (x2+2x – a) and q(x) = (x2 – x – 2) (x2 + 7x + b) have (x + 2) (x – 2) as their HCF.
Solution:
p(x) = (x2 + 5x + 6) (x2 + 2x – a)
= (x2 + 3x + 2x + 6) (x2 + 2x – a)
= x (x + 3) + 2 (x + 3) (x2 + 2x – a)
= (x + 2) (x + 3) (x2 + 2x – a)
Hence (x – 2) divides x2 + 2x – a
x2 + 2x – a = (x – 2) f(x)
Let x = 2
22 + 2(2) – a = 0
4 + 4 – a = 0
a = 8
q(x) = (x2 – x – 2) (x2 + 7x + b)
= (x2 – 2x + x – 2) (x2 + 7x + b)
= x (x – 2) + 1 (x – 2) (x2 + 7x + b)
= (x – 2) (x + 1) (x2 + 7x + b)
Hence (x + 2) divides x2 + 7x + b
x2 + 7x + b = (x + 2) g(x)
Let x = -2
(-2)2 + 7(-2) – b = 0
14 – 14 + b = 0
b = 10
- If the HCF of x2 + x – 12 and 2x2 – kx – 9 is (x – 1) find the value of k.
Solution:
Since x – k is the HCF of p(x) = x2 + x – 12 and 2×2 – kx – 9, it is a common factor of p(x) = 0 and q(x) = 0.
Put x = k
p(x) = k2 + k – 12 = 0
q(x) = 2k2 – k2 – 9 = 0
k2 – 9 = 0
k2 = 9
k = ± 3
Given p(x) = 0, k ≠ -3. Taking k = 3 we verify
p(x) = 3 + 3 – 12
= 9 + 3 – 12
= 12 – 12 =0
k = 3
1 thought on “HCF AND LCM – EXERCISE 3.3.3 – Class 9”
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