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Question 1. Factorise:
Solution:
Thus,
Question 2. Factorise: (x^{6} – y^{6})
Solution: x^{6} – y^{6 }= (x3)^{2} – (y3)^{2 }
= (x^{3} – y^{3})(x^{3} + y3) [∵ a^{2} – b^{2 }= (a + b)(a – b)]
= [(x – y)(x^{2} + xy + y^{2})][(x + y)(x^{2 }– xy + y^{2})] [∵ a^{3} + b^{3} = (a^{2} + b^{2} – ab)(a + b) and a^{3 }– b^{3} = (a^{2 }+ ab + b^{2})(a – b)]
= (x – y)(x + y)(x^{2 }+ xy + y^{2})(x^{2} – xy + y^{2})
Thus, x^{6 }– y^{6} = (x – y)(x + y)(x^{2} + xy + y^{2})(x^{2} – xy + y^{2})
Question 3. If the polynomials 2x^{3} + 3x^{2} – a and ax^{3} – 5x + 2 leave the same remainder when each is divided by x – 2, find the value of ‘a’
Solution: Let p(x) = 2x^{3} + 3x^{2} – a and f(x) = ax^{3 }– 5x + 2
when p(x) is divided by x – 2 then remainder = p(2) since p(2) = 2(2)^{3} + 3(2)^{2} – a = 2(8) + 3(4) – a = 16 + 12 – a ∴ Remainder = 28 – a  when f(x) is divided by x – 2, then remainder = f(2) since, f(2) = a(2)^{3} – 5(2) + 2 = a(8) – 10 + 2 = 8a – 8 ∴ Remainder = 8a – 8 
28 – a = 8a – 8
⇒ 8a + a = 28 + 8
⇒ 9a = 36
Thus , a = 4
Question 4. Find the values of ‘p’ and ‘q’, so that (x – 1) and (x + 2) are the factors of x^{3} + 10x^{2} + px + q.
Solution: Here f(x) = x^{3 }+ 10x^{2} + px + q
Since, x + 2 = 0 [∵ x + 2 is a factor of f(x)]
⇒ x= –2 If x + 2 is a factor f(x),
then f(–2) = 0 i.e. (–2)^{3 }+ 10(–2)^{2} + p(–2) + q = 0 [Factor theorem]
⇒ –8 + 40 + (–2p) + q = 0 ⇒ 32 – 2p + q = 0 ...(1)
⇒ 2p – q = 32 Also x – 1 = 0 ⇒ x = 1
If (x – 1) is a factor of f(x), then f(1) must be equal to 0. [Factor theorem]
i.e. (1)^{3 }+ 10(1)^{2} + p(1) + q = 0
⇒ 1 + 10 + p + q = 0
⇒ 11 + p + q = 0
⇒ p + q = –11 ...(2)
Now, adding (1) and (2), we get
Now we put p = 7 in (2), we have 7 + q = –11
⇒ q = –11 – 7 = –18
Thus, the required value of p and q are 7 and –18 respectively.
Question 5. If (x^{2} – 1) is a factor of the polynomial px^{4 }+ qx^{3} + rx^{2 }+ sx + t, then prove that p + r + t = q + s = 0.
Solution: We have f(x) = px^{4} + qx^{3} + rx^{2 }+ sx + t
Since, (x^{2} – 1) is a factor of f(x), [∵ x^{2} – 1 = (x + 1)(x – 1)]
then (x + 1) and (x – 1) are also factors of f(x).
∴ By factor theorem, we have f(1) = 0 and f(–1) = 0
For f(1) = 0, p(1)4 + q(1)^{3} + r(1)^{2} + s(1) + t = 0
⇒ p + q + r + s + t = 0 ...(1)
For f(–1) = 0, p(–1)^{4} + q(–1)^{3 }+ r(–1)^{2 }+ s(–1) + t = 0
⇒ p – q + r – s + t = 0 ...(2)
From (4) and (3), we get p + r + t = q + s = 0
Question 6. If x + y = 12 and xy = 27, find the value of x^{3} + y^{3}.
Solution: Since, (x + y)^{3} = x^{3} + y^{3} + 3xy (x + y)
∴ Substituting x + y = 12 and xy = 27,
we have: (12)^{3} =x^{3 }+ y^{3} + 3 (27) (12)
⇒ x^{3} + y^{3} = 81(12) – 123
= [92 – 122] × 12
= [(9 + 12) (9 – 12)] × 12
= 21 × 3 × 12 = 756
Question 7. If a + b + c = 5 and ab + bc + ca = 10, Then prove that a^{3 }+ b^{3} + c^{3} – 3abc = –25.
Solution: Since, a^{3} + b^{3} + c^{3 }– 3abc
= (a + b + c) (a^{2 }+ b^{2} + c^{2 }– ab – bc – ca)
∴ a^{3} + b^{3} + c^{3} – 3 abc = (a + b + c) [(a^{2} + b^{2} + c^{2} + 2ab+ 2bc+ 2cb) − 3ab − 3bc − 3ca]
= (a + b + c) [(a + b + c)2 − 3 (ab + bc + ca)]
= 5 [ 52 − 3(10)]
= 5[25 – 30]
= 5[–5] = –25
Question 8. If a, b, c are all nonzero and a + b + c = 0, prove that
Solution: Since, a + b + c = 0
∴ a^{3} + b^{3} + c^{3} = 3abc ..... (1)
Now, in = 3, we have
[Multiplying and dividing by ‘abc’]
..... (2)
From (1) and (2), we have
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