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**Question 1. Factorise: **

**Solution: **** **

Thus,

**Question 2. Factorise: (x ^{6} â€“ y^{6})**x

Solution:

= (x

= [(x â€“ y)(x

= (x â€“ y)(x + y)(x

Thus, x

** Question 3. If the polynomials 2x ^{3} + 3x^{2} â€“ a and ax^{3} â€“ 5x + 2 leave the same remainder when each is divided by x â€“ 2, find the value of â€˜aâ€™**Let p(x) = 2x

Solution:

when p(x) is divided by x â€“ 2 then remainder = p(2) since p(2) = 2(2) ^{3} + 3(2)^{2} â€“ a= 2(8) + 3(4) â€“ a = 16 + 12 â€“ a âˆ´ Remainder = 28 â€“ a | when f(x) is divided by x â€“ 2, then remainder = f(2) since, f(2) = a(2) ^{3} â€“ 5(2) + 2= a(8) â€“ 10 + 2 = 8a â€“ 8 âˆ´ Remainder = 8a â€“ 8 |

28 â€“ a = 8a â€“ 8

â‡’ 8a + a = 28 + 8

â‡’ 9a = 36

Thus , a = 4

**Question 4. Find the values of â€˜pâ€™ and â€˜qâ€™, so that (x â€“ 1) and (x + 2) are the factors of x ^{3} + 10x^{2} + px + q.** Here f(x) = x

Solution:

Since, x + 2 = 0 [âˆµ x + 2 is a factor of f(x)]

â‡’ x= â€“2 If x + 2 is a factor f(x),

then f(â€“2) = 0 i.e. (â€“2)

â‡’ â€“8 + 40 + (â€“2p) + q = 0 â‡’ 32 â€“ 2p + q = 0 ...(1)

â‡’ 2p â€“ q = 32 Also x â€“ 1 = 0 â‡’ x = 1

If (x â€“ 1) is a factor of f(x), then f(1) must be equal to 0. [Factor theorem]

i.e. (1)

â‡’ 1 + 10 + p + q = 0

â‡’ 11 + p + q = 0

â‡’ p + q = â€“11 ...(2)

Now, adding (1) and (2), we get

Now we put p = 7 in (2), we have 7 + q = â€“11

â‡’ q = â€“11 â€“ 7 = â€“18

Thus, the required value of p and q are 7 and â€“18 respectively.

**Question 5. If (x ^{2} â€“ 1) is a factor of the polynomial px^{4 }+ qx^{3} + rx^{2 }+ sx + t, then prove that p + r + t = q + s = 0.** We have f(x) = px

Solution:

Since, (x

then (x + 1) and (x â€“ 1) are also factors of f(x).

âˆ´ By factor theorem, we have f(1) = 0 and f(â€“1) = 0

For f(1) = 0, p(1)4 + q(1)

â‡’ p + q + r + s + t = 0 ...(1)

For f(â€“1) = 0, p(â€“1)

â‡’ p â€“ q + r â€“ 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}. **Since, (x + y)

Solution:

âˆ´ Substituting x + y = 12 and xy = 27,

we have: (12)

â‡’ x

= [92 â€“ 122] Ã— 12

= [(9 + 12) (9 â€“ 12)] Ã— 12

= 21 Ã— 3 Ã— 12 = 756

**Question 7. If a + b + c = 5 and ab + bc + ca = 10, Then prove that a ^{3 }+ b^{3} + c^{3} â€“ 3abc = â€“25. ** Since, a

Solution:

= (a + b + c) (a

âˆ´ a

= (a + b + c) [(a + b + c)2 âˆ’ 3 (ab + bc + ca)]

= 5 [ 52 âˆ’ 3(10)]

= 5[25 â€“ 30]

= 5[â€“5] = â€“25

Solution:

âˆ´ a

Now, in = 3, we have

[Multiplying and dividing by â€˜abcâ€™]

..... (2)

From (1) and (2), we have

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