 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
Pingback: IX – Table of Contents – Breath Math