Chapter 2: Polynomials and Factorisation (Exercise 2.5)

Chapter 2: Polynomials and Factorisation (Exercise 2.5)

Question 1:

Use suitable identities to find the following products:

1. $(x + 4)(x + 10)$

2. $(x + 8)(x - 10)$

3. $(3x + 4)(3x - 5)$

4. $\left(y^2 + \frac{3}{2}\right)\left(y^2 - \frac{3}{2}\right)$

5. $(3 - 2x)(3 + 2x)$

Solution:

1. $(x + 4)(x + 10)$:

   Using identity $(x + a)(x + b) = x^2 + (a + b)x + ab$ where $a = 4, b = 10$:

   $$= x^2 + (4 + 10)x + (4 \times 10) = x^2 + 14x + 40$$

2. $(x + 8)(x - 10)$:

   Using identity $(x + a)(x + b) = x^2 + (a + b)x + ab$ where $a = 8, b = -10$:

   $$= x^2 + (8 - 10)x + (8 \times -10) = x^2 - 2x - 80$$

3. $(3x + 4)(3x - 5)$:

   Using identity $(X + a)(X + b) = X^2 + (a + b)X + ab$ where $X = 3x, a = 4, b = -5$:

   $$= (3x)^2 + (4 - 5)(3x) + (4 \times -5) = 9x^2 - 3x - 20$$

4. $\left(y^2 + \frac{3}{2}\right)\left(y^2 - \frac{3}{2}\right)$:

   Using identity $(a + b)(a - b) = a^2 - b^2$ where $a = y^2, b = \frac{3}{2}$:

   $$= (y^2)^2 - \left(\frac{3}{2}\right)^2 = y^4 - \frac{9}{4}$$

5. $(3 - 2x)(3 + 2x)$:

   Using identity $(a - b)(a + b) = a^2 - b^2$ where $a = 3, b = 2x$:

   $$= (3)^2 - (2x)^2 = 9 - 4x^2$$

Question 2:

Evaluate the following products without multiplying directly:

1. $103 \times 107$

2. $95 \times 96$

3. $104 \times 96$

Solution:

1. $103 \times 107$:

   $$= (100 + 3)(100 + 7)$$

   Using $(x + a)(x + b) = x^2 + (a + b)x + ab$:

   $$= (100)^2 + (3 + 7)(100) + (3 \times 7) = 10000 + 1000 + 21 = 11021$$

2. $95 \times 96$:

   $$= (100 - 5)(100 - 4)$$

   Using $(x + a)(x + b) = x^2 + (a + b)x + ab$:

   $$= (100)^2 + (-5 - 4)(100) + (-5 \times -4) = 10000 - 900 + 20 = 9120$$

3. $104 \times 96$:

   $$= (100 + 4)(100 - 4)$$

   Using $(a + b)(a - b) = a^2 - b^2$:

   $$= (100)^2 - (4)^2 = 10000 - 16 = 9984$$

Question 3:

Factorise the following using appropriate identities:

1. $16x^2 + 24xy + 9y^2$

2. $4y^2 - 4y + 1$

3. $4x^2 - \frac{y^2}{25}$

Solution:

1. $16x^2 + 24xy + 9y^2$:

   Rewriting in the form $a^2 + 2ab + b^2$:

   $$= (4x)^2 + 2(4x)(3y) + (3y)^2 = (4x + 3y)^2 = (4x + 3y)(4x + 3y)$$

2. $4y^2 - 4y + 1$:

   Rewriting in the form $a^2 - 2ab + b^2$:

   $$= (2y)^2 - 2(2y)(1) + (1)^2 = (2y - 1)^2 = (2y - 1)(2y - 1)$$

3. $4x^2 - \frac{y^2}{25}$:

   Rewriting in the form $a^2 - b^2$:

   $$= (2x)^2 - \left(\frac{y}{5}\right)^2 = \left(2x + \frac{y}{5}\right)\left(2x - \frac{y}{5}\right)$$

Question 4:

Expand each of the following using suitable identities:

1. $(x + 2y + 4z)^2$

2. $(-2x + 3y + 2z)^2$

3. $(a - 2b - 3c)^2$

Solution:

Identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

1. $(x + 2y + 4z)^2$:

   $$= (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$$
   $$= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$$

2. $(-2x + 3y + 2z)^2$:

   $$= (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$$
   $$= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx$$

3. $(a - 2b - 3c)^2$:

   $$= (a)^2 + (-2b)^2 + (-3c)^2 + 2(a)(-2b) + 2(-2b)(-3c) + 2(-3c)(a)$$
   $$= a^2 + 4b^2 + 9c^2 - 4ab + 12bc - 6ca$$

Question 5:

Factorise $25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz$.

Solution:

Match with $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:

Since terms containing $x$ are negative ($-40xy$ and $-20xz$), take $x$ as negative:

Question 6:

Write the following cubes in expanded form:

1. $(2x + 1)^3$

2. $(2a - 3b)^3$

Solution:

1. $(2x + 1)^3$:

   Using $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:

   $$= (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)$$
   $$= 8x^3 + 1 + 6x(2x + 1) = 8x^3 + 12x^2 + 6x + 1$$

2. $(2a - 3b)^3$:

   Using $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$:

   $$= (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)$$
   $$= 8a^3 - 27b^3 - 18ab(2a - 3b) = 8a^3 - 27b^3 - 36a^2b + 54ab^2$$

Question 7:

Evaluate the following using suitable identities:

1. $(99)^3$

2. $(102)^3$

Solution:

1. $(99)^3 = (100 - 1)^3$:

   Using $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$:

   $$= (100)^3 - (1)^3 - 3(100)(1)(100 - 1)$$
   $$= 1000000 - 1 - 300(99) = 1000000 - 1 - 29700 = 970299$$

2. $(102)^3 = (100 + 2)^3$:

   Using $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:

   $$= (100)^3 + (2)^3 + 3(100)(2)(100 + 2)$$
   $$= 1000000 + 8 + 600(102) = 1000000 + 8 + 61200 = 1061208$$

Question 8:

Verify: $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

Solution:

Take Right Hand Side ($\text{RHS}$):

(Hence verified.)

Question 9:

Factorise $27y^3 + 125z^3$.

Solution:

Using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ where $a = 3y, b = 5z$:

Question 10:

Factorise $27x^3 + y^3 + z^3 - 9xyz$.

Solution:

Using identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$:

Question 11:

If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.

Solution:

We know that:

Substitute $x + y + z = 0$:

(Hence proved.)

Question 12:

Without actually calculating the cubes, find the value of $(-12)^3 + (7)^3 + (5)^3$.

Solution:

Let $x = -12, y = 7, z = 5$.

Checking sum:

Since $x + y + z = 0$, $x^3 + y^3 + z^3 = 3xyz$:

Question 13:

Give possible expressions for the length and breadth of the rectangle whose area is given by $25a^2 - 35a + 12$.

Solution:

Factorising by splitting the middle term (Product = $25 \times 12 = 300$, Sum = $-35$ $\rightarrow$ Factors $-20, -15$):

• Possible Length: $(5a - 3)$

• Possible Breadth: $(5a - 4)$

Question 14:

What are the possible expressions for the dimensions of the cuboid whose volume is given by $3x^2 - 12x$?

Solution:

Factoring out common terms:

• Dimensions: $\text{Length} = 3$, $\text{Breadth} = x$, $\text{Height} = (x - 4)$