HCF AND LCM – EXERCISE 3.3.3 – Class 9

1. 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)

---

2. 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)

---

3. 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)

---

4. 4. 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)

---

5. 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)

---

6. 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

---

7. 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

---

8. 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

---

9. 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

---