**Question 1**:

**Without actual division, show that (x**^{3}** – 3x**^{2}** – 13x + 15) is exactly divisible by (x**^{2}** + 2x – 3).**

Let f(x) = x^{3} – 3x^{2} – 13x + 15

Now, x^{2} + 2x – 3 = x^{2} + 3x – x – 3

= x (x + 3) – 1 (x + 3)

= (x + 3) (x – 1)

Thus, f(x) will be exactly divisible by x2 + 2x – 3 = (x + 3) (x – 1)

if (x + 3) and (x – 1) are both factors of f(x), so by factor theorem,

we should have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)^{3} – 3 (-3)^{2} – 13 (-3) + 15

= -27 – 3 x 9 + 39 + 15

= -27 – 27 + 39 + 15

= -54 + 54 = 0

And, f(1) = 1^{3} – 3 X 1^{2} – 13 x 1 + 15

= 1 – 3 – 13 + 15

= 16 – 16 = 0

f(-3) = 0 and f(1) = 0

So, x^{2} + 2x – 3 divides f(x) exactly.

**Question 2**:

**If (x**^{3}** + ax**^{2}** + bx + 6) has (x-2) as a factor and leaves a remainder 3 when divided by (x-3), find the values of a and b.**

Let f(x) = (x^{3} + ax^{2} + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 3^{3} + a x 2^{2} + b x 3 + 6 = 3

= 27 + 9a + 3b + 6 = 3

= 9a + 3b + 33 = 3

= 9a + 3b = 3-33

= 9a + 3b = -30

= 3a + b = -10 ….(i)

Given that (x – 2) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0

. f(2) = 2^{3} + a x 2^{2} + b x 2 + 6 = 0

= 8 + 4a+ 2b + 6 = 0

= 4a + 2b = -14

= 2a + b = -7 ….(ii)

Subtracting (ii) from (i), we get,

a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

a = -3 and b = -1.

**Question 3**:

**9x**^{2}** + 12xy **

9x^{2} + 12xy = 3x (3x + 4y)

**Question 4**:

**18 x**^{2}**y – 24xyz **

18 x^{2}y – 24xyz = 6xy (3x – 4z)

**Question 5: **

**27 a**^{3}** b**^{3}** – 45 a**^{4}** b**^{2}

27 a^{3} b^{3} – 45 a^{4} b^{2} = 9a^{3} b^{2} (3b – 5a)

**Question 6:**

**2a (x + y) – 3b (x + y) **

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)

**Question 7: **

**2x (p**^{2}** + q**^{2}**) +4y (p**^{2}** + q**^{2}**) **

2x (p^{2} + q^{2}) +4y (p^{2} + q^{2})

= (2x + 4y) (p^{2} + q^{2})

= 2(x+ 2y) (p^{2} + q^{2})

**Question 8: **

**x (a – 5) + y (5 – a)**

x (a – 5) + y (5 – a)

= x (a – 5) + y (-1) (a – 5)

= (x – y) (a – 5)

**Question 9:**

**4 (a + b) – 6 (a + b)**^{2}

4 (a + b) – 6 (a + b)^{2}

= (a + b) [4 – 6 (a + b)]

= 2 (a + b) (2 – 3a – 3b)

= 2 (a + b) (2 – 3a – 3b)

**Question 10:**

**8 (3a – 2b)**^{2}** – 10 (3a – 2b) **

8 (3a – 2b)^{2} – 10 (3a – 2b)

= (3a – 2b) [8(3a – 2b) – 10]

= (3a – 2b) 2[4 (3a – 2b) – 5]

= 2 (3a – 2b) (12 a – 8b – 5)

**Question 11:**

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

x (x + y)^{3} – 3 x^{2} y (x + y)

= x (x + y) [(x + y)^{2} – 3xy]

= x (x + y) (x^{2} + y^{2} + 2xy – 3xy)

= x (x + y) (x^{2} + y^{2} – xy)

**Question 12:**

**X**^{3}** + 2x**^{2}** + 5x + 10**

X^{3} + 2x^{2} + 5x + 10

= x^{2} (X + 2) + 5 (x + 2)

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

**Question 13:**

**X**^{2}** + xy – 2xz – 2yz **

X^{2} + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)

**Question 14:**

**a**^{3}** b – a**^{2}** b + 5ab – 5b **

a^{3} b – a^{2} b + 5ab – 5b

= a^{2} b (a – 1) + 5b (a – 1)

= (a – 1) (a^{2} b + 5b)

= (a – 1) b (a^{2} + 5)

= b (a – 1) (a^{2} + 5)

**Question 15:**

**8 – 4a – 2a**^{3}** + a4 **

8 – 4a – 2a^{3} + a4

= 4(2 – a) – a^{3} (2 – a)

= (2 – a) (4 – a^{3})

**Question 16: **

**X**^{3}** – 2x**^{2}** y + 3xy**^{2}** – 6y**^{3}

X^{3} – 2x^{2} y + 3xy^{2} – 6y^{3}

= x^{2} (x – 2y) + 3y^{2} (x – 2y)

= (x – 2y) (x^{2} + 3y^{2})

**Question 17: **

**px + pq – 5q – 5x **

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)

**Question 18**:

**X**^{2}** – xy + y – x **

X^{2} – xy + y – x

= x (x – y) – 1 (x – y)

= (x – y) (x – 1)

**Question 19:**

**(3a – 1)**^{2}** – 6a + 2 **

(3a – 1)^{2} – 6a + 2

= (3a – 1)^{2} – 2 (3a – 1)

= (3a – 1) [(3a – 1) – 2]

= (3a – 1) (3a – 3)

= 3(3a – 1) (a – 1)

**Question 20: **

**(2x – 3)**^{2}** – 8x + 12**

(2x – 3)^{2} – 8x + 12

= (2x – 3)^{2} – 4 (2x – 3)

= (2x – 3) (2x – 3 – 4)

= (2x – 3) (2x – 7)

**Question 21:**

**a**^{3}** + a – 3a**^{2}** – 3 **

a^{3} + a – 3a^{2} – 3

= a(a^{2} + 1) – 3 (a^{2} + 1)

= (a – 3) (a^{2} + 1)

**Question 22: **

**3ax – 6ay – 8by + 4bx **

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)

**Question 23:**

**abx**^{2}** + a**^{2}** x + b**^{2}** x +ab **

abx^{2} + a^{2} x + b^{2} x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

**Question 24: **

**X**^{3}** – x**^{2}** + ax + x – a – 1 **

X^{3} – x^{2} + ax + x – a – 1

= x^{3} – x^{2} + ax – a + x – 1

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

= (x – 1) (x^{2} + a + 1)

**Question 25:**

**2x + 4y – 8xy – 1 **

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)

**Question 26: **

**ab (x**^{2}** + y**^{2}**) – xy (a**^{2}** + b**^{2}**)**

ab (x^{2} + y^{2}) – xy (a^{2} + b^{2})

= abx^{2} + aby^{2} – a^{2} xy – b^{2} xy

= abx^{2} – a^{2} xy + aby^{2} – b^{2} xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)

**Question 27:**

**a**^{2}** + ab (b + 1) + b**^{3}

a^{2} + ab (b + 1) + b^{3}

= a^{2} + ab^{2} + ab + b^{3}

= a^{2} + ab + ab^{2} + b^{3}

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

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

**Question 28: **

**a**^{3}**+ab(1-2a)-2b**^{2}

a^{3}+ab(1-2a)-2b^{2}

= a^{3} + ab – 2a^{2} b – 2b^{2}

= a (a^{2} + b) – 2b (a^{2} + b)

= (a^{2} + b) (a – 2b)

**Question 29:**

**2a**^{2}** + bc – 2ab – ac **

2a^{2} + bc – 2ab – ac

= 2a^{2} – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)

**Question 30: **

**(ax + by)**^{2}** + (bx – ay)**^{2}

(ax + by)^{2} + (bx – ay)^{2}

= a^{2} x^{2} + b^{2} y^{2} + 2abxy + b^{2} x^{2} + a^{2} y^{2} – 2abxy

= a^{2} x^{2} + b^{2} y^{2} + b^{2} x^{2} + a^{2} y^{2}

= a^{2} x^{2} + b^{2} x^{2}+ b^{2} y^{2} + a^{2} y^{2}

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

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

**Question 31: **

**a (a + b – c) – bc**

a (a + b – c) – bc

= a2 + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)

**Question 32: **

**a(a – 2b – c) + 2bc **

a(a – 2b – c) + 2bc

= a^{2} – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)

**Question 33: **

**a**^{2}** x**^{2}** + (ax**^{2}** + 1)x + a **

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

= a^{2} x^{2} + ax^{3} + x + a

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

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

**Question 34: **

**ab (x**^{2}** + 1) + x (a**^{2}** + b**^{2}**) **

ab (x^{2} + 1) + x (a^{2} + b^{2})

= abx^{2} + ab + a^{2} x + b^{2} x

= abx^{2} + a^{2} x + ab + b^{2} x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

**Question 35:**

**X**^{2}**– (a + b) x + ab**

X^{2}– (a + b) x + ab

= x2 – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)

**Question 36:**

**Question 37**:

**25x**^{2}** – 64y**^{2}

25x^{2} – 64y^{2}

= (5x)^{2} – (8y)^{2}

= (5x + 8y) (5x – 8y)

[Therefore, a^{2}-b

^{2}= (a + b) (a – b)]

**Question 38**:

100 – 9x^{2}

100 – 9x^{2}

= (10)^{2} – (3x)^{2}

= (10 + 3x) (10 – 3x)

[Therefore, a^{2}– b

^{2}= (a + b) (a – b)]

**Question 39**:

**5x**^{2}** – 7y**^{2}

5x^{2} – 7y^{2}

^{2}– b

^{2}= (a + b) (a – b)]

**Question 40:**

**(3x + 5y)**^{2}** – 4z**^{2}

(3x + 5y)^{2} – 4z^{2}

= (3x + 5y)^{2} – (2z)^{2}

= (3x + 5y + 2z) (3x + 5y – 2z)

[Therefore, a^{2}– b

^{2}= (a + b) (a-b)]

**Question 41:**

**150 – 6x**^{2}

150 – 6x^{2}

= 6 (25 – x^{2})

= 6 ( 5^{2 }– x^{2})

= 6 (5 + x) (5 – x)

[Therefore, a^{2}– b

^{2}= (a + b) (a-b)]

**Question 42**:

**20x**^{2}** – 45**

20x^{2} – 45

= 5(4x^{2} – 9)

= 5 [(2x)^{2} – (3)^{2}]

= 5 (2x + 3) (2x – 3)

[Therefore, a^{2}– b

^{2}= (a + b) (a-b)]

**Question 43**:

**3x**^{3}** – 48x**

3x^{3} – 48x

= 3x (x^{2} – 16)

= 3x [(x)^{2} – (4)^{2}]

= 3x (x + 4) (x – 4)

[Therefore, a^{2}– b

^{2}= (a + b) (a-b)]

**Question** **44**:

**2 – 50x**^{2}

2 – 50x^{2}

= 2 (1 – 25x^{2})

= 2 [(1)^{2} – (5x)^{2}]

= 2 (1 + 5x) (1 – 5x)

[Therefore a^{2}– b

^{2}= (a + b) (a-b)]

**Question 45**:

**27a**^{2}**– 48b**^{2}

27a^{2}– 48b^{2}

= 3 (9a^{2} – 16b^{2})

= 3 [(3a)^{2} – (4b)^{2}]

= 3(3a + 4b) (3a – 4b)

[Therefore, a^{2}– b

^{2}= (a + b) (a-b)]