**Find the HCF of each of the following pairs of polynomials**

**(i) x**^{2}**+ 2x – 15 and x**^{2}**– 7x + 12 **

Solution:

x^{2} + 2x – 15 & x^{2} – 7x + 12

= x^{2} + 5x – 3x – 15 = x^{2} – 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) m**^{2}**– 3m + 2 and m**^{2}**+ m – 6 **

Solution:

m^{2} – 3m + 2

= m^{2} – 2m – m + 2

= m (m – 2) – 1 (m – 2)

= (m – 2) (m – 1)

m^{2} + m – 6

= m^{2} + 3m – 2m – 6

= m (m + 3) – 2 (m + 3)

= (m + 3) (m – 2)

**HCF = (m – 2)**

**(iii) x**^{2}**– xy – 2y**^{2 }**and x**^{2 }**+ 3xy + 2y**^{2 }

x^{2} – xy – 2y^{2} = x^{2} – 2xy + xy – 2y^{2}

= x (x – 2y) + y (x – 2y)

= (x – 2y) (x + y)

x^{2} + 3xy + 2y^{2}

= x^{2} + 3xy + xy + 2y^{2 }

= x (x + 2y) + y (x + 2y)

= (x + 2y) (x + y)

**HCF = (x + y) **

**(iv) x**^{2}**– 2x + 1 and x**^{2}**– 3x + 2 **

x^{2} – 2x + 1 = (x – 1)^{2}

x^{2} – 3x + 2 = x^{2} – 2x – x + 2

= x (x – 2) – 1 (x – 2)

= (x – 2) (x – 1)

**HCF = (X – 1)**

**Find the HCF of p(x) = (x**^{3}**-27) (x**^{2}**– 3x +2) and**

**q(x) =(x**^{2 }**+ 3x + 9) (x**^{2}**– 5x +6) **

Solution:

p(x) = (x^{3} -27) (x^{2} – 3x +2)

(x^{3} -27) = x^{3 }– 9^{3} = (x – 3) (x^{2} + 3x + 3^{2})

= (x – 3) (x^{2} + 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) = (x^{2} + 3x + 9) (x^{2} – 5x +6)

(x^{2} – 5x +6) = x – 3x – 2 x + 6

=x (x – 3) – 2 (x – 3)

= (x – 3) (x – 2)

q(x) = (x – 3) (x – 2) (x^{2} + 3x + 9)

**HCF = (x – 3) (x – 2) (x**^{2}**+ 3x + 9)**

**Find the HCF of f(x) = x**^{3}**+ x**^{2}**– x – 1 and g(x) =****x**^{3}**+ x**+^{2}**x + 1**

Solution:

f(x) = x^{3} + x^{2} – x – 1

= x^{3} (x + 1) – 1 (x + 1)

= (x^{2} – 1) (x + 1)

= (x + 1) (x – 1) (x + 1)

= (x + 1)^{2} (x – 1)

g(x) = x^{3} + x^{2} + x + 1

= x^{2} (x + 1) + 1 (x + 1)

= (x^{2} – 1) (x + 1)

**HCF = (x + 1)**

**Find the HCF of p(x) = x**^{4 }– 2x^{3 }– 15x^{2}and q(x) = x^{3}– 9x

p(x) = x

^{4}– 2x^{3}– 15x^{2}= x

^{2}(x^{2}– 2x – 15)= x

^{2}(x^{2}-5x + 3x – 15)= x

^{2}[x (x – 5) + 3(x – 5)]= x

^{2}(x + 3) (x – 5)q(x) = x

^{3}– 9x

= x (x^{2} – 9)

= x (x^{2} – 3^{2})

= 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) a**^{4}**b – ab**^{4}**, a**^{4}**b**^{2}**– a**^{2}**b**^{4}**and a**^{2}**b**^{2}**(a**^{4}**– b**^{4}**) **

a^{4}b – ab^{4} = ab (a^{3} – b^{3})

= ab (a – b) (a^{2} + 2ab +b^{2})

a^{4}b^{2} – a^{2}b^{4} = a^{2} b^{2} (a^{2} – b^{2})

= a^{2} b^{2} (a – b) (a + b)

a^{2 }b^{2} (a^{4} – b^{4}) = a^{2} b^{2} (a^{2 }+ b^{2}) (a^{2} – b^{2})

= a^{2} b^{2} (a^{2} + b^{2}) (a – b) (a + b)

**HCF = ab (a – b)**

**(ii) 6(x**^{2}**+ 10x + 24), 4(x**^{2 }**– x – 20) and 8(x**^{2}**+ 3x – 4) **

Solution:

6(x^{2} + 10x + 24) = 6(x^{2} + 6x + 4 x + 247)

= 6 [x (x + 6) + 4 (x + 6)]

= 2 x 3 (x + 6) (x + 4)

4(x^{2 }– x – 20) = 4 (x – 5x + 4x – 20)

= 4 [x (x – 5) + 4 (x – 5)]

= 2^{2} (x – 5) (x + 4)

8(x^{2} + 3x – 4) = 2^{3} (x^{2} + 4x – x – 4)

= 2^{3} [x (x + 4) – 1 (x + 4)

= 2^{3} (x + 4) (x – 1)

**HCF = 2 (x + 4) **

**(iii) a**^{2}**(2x**^{2}**+ 5x + 2), a**^{2}**b (3x**^{2}**+ 8x +4)and ab**^{2}**(2x**^{2}**+ 3x – 2) **

Solution:

a^{2} (2x^{2} + 5x + 2) = a^{2} (2x^{2} + 4x + x + 2)

= a^{2} [2x (x + 2) + 1 (x + 2)]

= a^{2} (2x + 1) (x + 2)

a^{2}b (3x^{2} + 8x +4) = a^{2}b (3x^{2} + 6x + 2x +4)

= a^{2}b [3x(x + 2) + 2 (x + 2)]

= a^{2}b (x + 2) (3x + 2)

ab^{2} (2x^{2} + 3x – 2) = ab^{2} (2x^{2} + 4x – x – 2)

= ab^{2} [2x (x + 2) – 1 (x + 2)]

= ab^{2} (2x – 1) (x + 2)

**HCF is a (x + 2) HCF = a (x + 2) **

**If the HCF of the polynomials p(x) = (x + 1) (x**^{2}**+ ax + 4) and**

**q(x) = (x + 4) (x²**** ****+ bx +2) is h(x) = x**^{2}**+ 5x + 4, find the values of a and b. **

Solution:

p(x) = (x + 1) (x^{2} + ax + 4)

q(x) = (x + 4) (x^{2} + bx +2)

h(x) = x^{2} + 5x + 4

= x^{2} + 4x + x + 4

= x (x + 4) + 1(x + 4)

= (x + 4) (x +1)

Hence (x + 1) divides x^{2} + bx + 2

x^{2} + bx + 2 = (x + 1) f(x)

Let x = –1

(–1)^{2} + b (–1) + 2 = 0

–b + 3 = 0

**b = 3**

Also (x + 4) divides x^{2} + ax + 4

x^{2} + 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) (2x**^{2}**– ax + 2) and**

**q(x) = (x + 4) (x**^{2}**– bx – 6) is h(x) = x**^{2}**– 5x + 6, find the values of a and b. **

Solution:

h(x) = x^{2} – 5x + 6

= x^{2} – 3x – 2x + 6

= x (x – 3) – 2 (x – 3)

= (x – 3) (x – 2)

p(x) = (x – 3) (2x^{2} – ax + 2)

Hence (x – 2) divides 2x^{2} – ax + 2

2x^{2} – 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) (x^{2 }– bx – 6)

Hence (x – 3) divides (x^{2} – bx – 6)

(x^{2} – bx – 6)= (x – 3) g(x)

Let x = 3

3^{3} – b (3) – 6 = 0

9 – 3b – 6 = 0

3b = 3

**b = 1 **

**For what values of a and b the polynomials p(x) = (x**^{2}**+5x+6) (x**^{2}**+2x – a) and q(x) = (x**^{2}**– x – 2) (x**^{2}**+ 7x + b) have (x + 2) (x – 2) as their HCF.**

Solution:

p(x) = (x^{2} + 5x + 6) (x^{2} + 2x – a)

= (x^{2} + 3x + 2x + 6) (x^{2} + 2x – a)

= x (x + 3) + 2 (x + 3) (x^{2} + 2x – a)

= (x + 2) (x + 3) (x^{2 }+ 2x – a)

Hence (x – 2) divides x^{2} + 2x – a

x^{2} + 2x – a = (x – 2) f(x)

Let x = 2

2^{2} + 2(2) – a = 0

4 + 4 – a = 0

**a = 8 **

q(x) = (x^{2} – x – 2) (x^{2} + 7x + b)

= (x^{2} – 2x + x – 2) (x^{2} + 7x + b)

= x (x – 2) + 1 (x – 2) (x^{2} + 7x + b)

= (x – 2) (x + 1) (x^{2} + 7x + b)

Hence (x + 2) divides x^{2} + 7x + b

x^{2} + 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 x**^{2}**+ x – 12 and 2x**^{2 }**– kx – 9 is (x – 1) find the value of k.**

Solution:

Since x – k is the HCF of p(x) = x^{2} + 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) = k^{2} + k – 12 = 0

q(x) = 2k^{2 }– k^{2} – 9 = 0

k^{2} – 9 = 0

k^{2} = 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**

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