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**Q.1. Write the numerical co-efficient and degree of each term of: ****Sol.**

**Q.2. Find the remainder when x ^{3} – ax^{2} + 4x – a is divided by (x – a).**

Sol.

p(x) = x

(x – a) = 0

► x = a

∴ p(a) = (a)

∴ The required remainder = 3a.

Find the value of k.

Sol.

Here, p(x) = kx

Since, Divisor = x + 3

∴ x + 3 = 0

► x = –3

∴ p(–3) = 7

► k(–3)

► –27k + 81 – 12 – 8 = 7

► –27k = 7 – 81 + 12 + 8

► –27k = 27 – 81

► –27k = –54

► k= -(54/27) = 2

Thus, the required value of k = 2.

(b) For what value of ‘m’ is x

If p(x) is exactly divisible by x – 2, then p(2) = 0

i.e. (2)

► 4 + 8 – 2k + 2 = 0

► 14 – 2k = 0

► 2k = 14

► k= 14/2 = 7

Thus, the required value of k is 7.

**(b)** Here, p(x) = x^{3} – 2mx^{2} + 16

∴ p(–2) = (–2)^{3} –2(–2)^{2}m + 16

= –8 –8m + 16

= –8m + 8

Since, p(x) is divisible by x + 2

∴ p(–2) = 0

or –8m + 8 = 0

► m = 1**Q.5. Factorize x ^{2} – x – 12.Sol.**

We have x

= x^{2 }– 4x + 3x – 12

= x(x – 4) + 3(x – 4)

= (x – 4)(x + 3)

Thus, x^{2 }– x – 12 = (x – 4)(x + 3)**Q.6. If x + (1/2)x = 5, then find the value of x ^{2} + **

**Sol.**

We have x + (1/2)x = 5

__Squaring both sides, we get:__

►

►

►

Thus, the required value of **Q.7. Check whether (x – 1) is a factor of the polynomial x ^{3} – 27x^{2} + 8x + 18.**

Sol.

Here, p(x) = x

For x – 1 = 0

► x = 1

∴ p(1) = (1)

= 1 – 27 + 8 + 18

= 27 – 27

= 0

Since, p(1) = 0

∴ (x – 1) is a factor of p(x).

Thus, (x – 1) is a factor of x

**Sol. **

Here f(x) =

Divisor = x+(2/3)

since, x+(2/3) = 0 ⇒ x = -2/3

∴ Remainder = f (-2/3)

i.e. Remainder

Thus, the required remainder = (10/27)**Q.9. Find the value of k, if (x – k) is a factor of x ^{6} – kx^{5} + x^{4} – kx^{3} + 3x – k + 4.**

Sol.

Here, p(x) = x

If (x – k) is a factor of p(x), then p(k) = 0

i.e (k)

► k

► 2k + 4 = 0

► 2k = – 4

► k = (-4/2) = –2

Thus, the required value of k is –2.

**Q.10. Factorize: 9a ^{2} – 9b^{2} + 6a + 1**

Sol.

9a

= [9a

= [(3a)

= (3a + 1)

= [(3a + 1) + 3b][(3a + 1) – 3b] {using x

= (3a + 1 + 3b)(3a + 1 – 3b)

Sol.

x^{3} + y^{3 }– 12xy + 64

= (x)^{3} + (y)^{3} + (4)^{3 }– 3(x)(y)(4)

= [x^{2} + y^{2} + 42 – xy – y. 4 – 4 . x](x + y + 4)

= [x^{2} + y^{2} + 16 – xy – 4y – 4x][x + y + 4] ...(1)

Since, x + y = –4 ∴ x + y + 4 = 0 ...(2)

From (1) and (2), we have x^{3} + y^{3 }– 12xy + 64

= [x^{2 }+ y^{2} + 16 – xy – 4y – 4x][0] = 0

Thus, x^{3} + y^{3} – 12xy + 64 = 0.

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