ICSE Class 8 Maths Algebra Chapter 16 Special Products and Expansions PDF Download

Special Products And Expansions

Special Products

In the previous class we used geometry to show that:

\[(a + b) (a - b) = a^2 - b^2\]

Multiplying the two binomials we have

\[(a + b) (a - b) = a^2 - ab + ab - b^2 = a^2 - b^2\]

Thus, we observe a pattern that emerges in the special product of the two binomials as:

\[(term 1 + term 2) (term 1 - term 2) = (term 1)^2 - (term 2)^2\]

Example 1

Apply the identity \[(a + b) (a - b) = a^2 - b^2\] to find the value of 9015 × 8985.

\[9015 \times 8985 = (9000 + 15)(9000 - 15) = 9000^2 - 15^2 = 80999775\]

Let us now study special products involving three terms. Let x, a, and b be three terms in two binomials. Then we have

1. \[(x + a) (x + b) = x^2 + xb + ax + ab = x^2 + x(a + b) + ab\]

2. \[(x - a) (x - b) = x^2 - bx - ax + ab = x^2 - x(a + b) + ab\]

3. \[(x + a) (x - b) = x^2 - bx + ax - ab = x^2 + x(a - b) - ab\]

4. \[(x - a) (x + b) = x^2 + bx - ax - ab = x^2 + x(b - a) - ab\]

Example 2

Find the product of \[(x + 3) (x + 8)\]

Applying \[(x + a) (x + b) = x^2 + x(a + b) + ab\]

\[(x + 3) (x + 8) = x^2 + x(3 + 8) + (3 \times 8) = x^2 + 11x + 24\]

Example 3

Find the product of \[\left(\frac{p}{5} + \frac{1}{2}\right)\left(\frac{p}{5} - \frac{1}{3}\right)\]

Let \[\frac{p}{5} = x, \frac{1}{2} = a\], and \[\frac{1}{3} = b\]. Then we observe a pattern resembling \[(x + a) (x - b)\] and we can write the product without having to actually multiply the binomials.

Thus, \[\left(\frac{p}{5} + \frac{1}{2}\right)\left(\frac{p}{5} - \frac{1}{3}\right) = \left(\frac{p}{5}\right)^2 + \frac{p}{5}\left(\frac{1}{2} - \frac{1}{3}\right) - \frac{1}{2} \times \frac{1}{3} = \frac{p^2}{25} + \frac{p}{30} - \frac{1}{6}\]

Example 4

\[(3x^3 + 10) (3x^3 + 15)\]

\[= (3x^3)^2 + 3x^3(10 + 15) + 10 \times 15\]

\[[applying (x + a)(x + b) = x^2 + x(a + b) + ab]\]

\[= 9x^6 + 75x^3 + 150\]

Example 5

Find the continued product of \[(x - 1) (x + 1) (x^2 + 1) (x^4 + 1) (x^8 + 1)\]

\[= (x^2 - 1) (x^2 + 1) (x^4 + 1) (x^8 + 1)\]

\[[applying (a - b) (a + b) = a^2 - b^2]\]

\[= (x^4 - 1) (x^4 + 1) (x^8 + 1)\]

\[= (x^8 - 1) (x^8 + 1) = x^{16} - 1\]

Expansions

The number 121 is written in exponential form as \[11^2\], which can be written in expanded form as 11 × 11.

Similarly, the algebraic expression \[(a + b)^2\] is written in expanded form as:

\[(a + b)(a + b) = a^2 + 2ab + b^2\]

In the previous class we used geometry to show that:

\[(a + b)^2 = a^2 + 2ab + b^2\]

and \[(a - b)^2 = a^2 - 2ab + b^2\]

Observe the pattern of both expansions:

\[(term 1 + term 2)^2 = (term 1)^2 + (2 \times term 1 \times term 2) + (term 2)^2\]

and \[(term 1 - term 2)^2 = (term 1)^2 - (2 \times term 1 \times term 2) + (term 2)^2\]

Example 6

Solve \[\left(3x + \frac{y}{6}\right)^2\]

Let \[3x = a\] and \[\frac{y}{6} = b\]. Observe that we obtain a pattern resembling \[(a + b)^2\] and we can expand the expression without having to multiply the binomial with itself.

Thus, \[\left(3x + \frac{y}{6}\right)^2 = (term 1)^2 + (2 \times term 1 \times term 2) + (term 2)^2\]

\[= (3x)^2 + \left(2 \times 3x \times \frac{y}{6}\right) + \left(\frac{y}{6}\right)^2\]

\[= 9x^2 + xy + \frac{y^2}{36}\]

Example 7

What should be added to \[49x^2 - 74xy + 25y^2\] to make it a perfect square?

The expanded form of a perfect square in the form \[(term 1 - term 2)^2\] would be:

\[(term 1)^2 - (2 \times term 1 \times term 2) + (term 2)^2\]

In the expression \[49x^2 - 74xy + 25y^2\], the first and last terms are squares of 7x and 5y respectively, but if 7x - 5y were to be squared, the middle term of the expanded form would be \[-(2 \times 7x \times 5y) = -70xy\], whereas in the given expression the middle term is -74xy.

Thus, if \[-70xy - (-74xy) = + 4xy\] were to be added to \[49x^2 - 74xy + 25y^2\], the sum \[49x^2 - 70xy + 25y^2\] would be a perfect square.

Example 8

Expand \[(a + b - c)^2\]

If \[(a + b)\] is considered as a single term then the expression can be written as:

\[\{(a + b) - c\}^2 = (a + b)^2 - \{2 \times (a + b) \times c\} + c^2\]

\[= a^2 + 2ab + b^2 - (2ac + 2bc) + c^2\]

\[= a^2 + b^2 + c^2 - 2ab - 2ac - 2bc + c^2\]

\[= a^2 + b^2 + c^2 + 2(ab + ac + bc)\]

Example 9

Given that \[x + \frac{1}{x} = 4\], find the value of \[x^4 + \frac{1}{x^4}\]

Squaring both the left hand side and right hand side of the given statement does not change the equation.

So, \[\left(x + \frac{1}{x}\right)^2 = 4^2\]

\[\Rightarrow x^2 + \left(2 \times x \times \frac{1}{x}\right) + \frac{1^2}{x^2} = 16\]

\[\Rightarrow x^2 + \frac{1}{x^2} = 16 - 2 = 14\]

\[\Rightarrow \left(x^2 + \frac{1}{x^2}\right)^2 = 14^2\]

\[\Rightarrow x^4 + 2 + \frac{1}{x^4} = 196\]

\[\Rightarrow x^4 + \frac{1}{x^4} = 196 - 2 = 194\]

Example 10

Given \[x^2 + \frac{1}{x^2} = 51\], find the value of \[x - \frac{1}{x}\]

Adding 2 to \[x^2 + \frac{1}{x^2}\] gives us \[x^2 + 2 + \frac{1}{x^2}\] which is the expanded form of \[\left(x + \frac{1}{x}\right)^2\]

Subtracting 2 from \[x^2 + \frac{1}{x^2}\] gives us \[x^2 - 2 + \frac{1}{x^2}\] which is the expanded form of \[\left(x - \frac{1}{x}\right)^2\]

Thus, subtracting 2 from both sides of the given statement,

\[x^2 - 2 + \frac{1}{x^2} = 51 - 2\]

\[\Rightarrow \left(x - \frac{1}{x}\right)^2 = 49 \Rightarrow x - \frac{1}{x} = \sqrt{49} = 7\]

Example 11

Given that \[x^2 + y^2 = 74\] and \[xy = 35\], find \[x + y\] and \[x - y\]

\[(x + y)^2 = x^2 + 2xy + y^2\]

\[\Rightarrow (x + y)^2 = x^2 + y^2 + 2xy\]

\[\Rightarrow (x + y)^2 = 74 + (2 \times 35)\]

\[\Rightarrow (x + y)^2 = 144\]

\[\Rightarrow x + y = \sqrt{144} = 12\]

\[(x - y)^2 = x^2 - 2xy + y^2\]

\[\Rightarrow (x - y)^2 = x^2 + y^2 - 2xy\]

\[\Rightarrow (x - y)^2 = 74 - (2 \times 35)\]

\[\Rightarrow (x - y)^2 = 4\]

\[\Rightarrow x - y = \sqrt{4} = 2\]

Having learnt and applied the two expansions \[(a + b)^2\] and \[(a - b)^2\], let us study some more expansions.

1. \[(a + b)^3 = (a + b) (a + b) (a + b)\]

\[= (a^2 + 2ab + b^2) (a + b)\]

\[= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3\]

\[= a^3 + 3a^2b + 3ab^2 + b^3\]

\[= a^3 + 3ab (a + b) + b^3\]

Pattern: \[(term 1 + term 2)^3 = (term 1)^3 + 3 \times term 1 \times term 2 (term 1 + term 2) + (term 2)^3\]

\[= sum of cubes of terms + three times the product of their sum and product\]

2. \[(a - b)^3 = (a - b) (a - b) (a - b)\]

\[= (a^2 - 2ab + b^2) (a - b)\]

\[= a^3 - a^2b - 2a^2b + 2ab^2 + ab^2 - b^3\]

\[= a^3 - 3a^2b + 3ab^2 - b^3\]

\[= a^3 - 3ab (a - b) - b^3\]

Pattern: \[(term 1 - term 2)^3 = (term 1)^3 - 3 \times term 1 \times term 2 (term 1 - term 2) - (term 2)^3\]

\[= difference of cubes of terms - three times the product of their difference and product\]

3. \[(a + b + c)^2 = (a + b + c)(a + b + c)\]

\[= a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2\]

\[= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc\]

\[= a^2 + b^2 + c^2 + 2(ab + ac + bc)\]

Pattern: \[(term 1 + term 2 + term 3)^2 = (term 1)^2 + (term 2)^2 + (term 3)^2 + 2(term 1 \times term 2 + term 1 \times term 3 + term 2 \times term 3)\]

\[= sum of squares of the three terms + two times the sum of their products in pairs.\]

Example 12

Expand \[(2x + 3y)^3\]

Following the pattern of the cube of the sum of two terms,

\[(2x + 3y)^3 = (2x)^3 + (3y)^3 + \{3 \times 2x \times 3y (2x + 3y)\}\]

\[= 8x^3 + 27y^3 + \{18xy (2x + 3y)\}\]

\[= 8x^3 + 27y^3 + \{36x^2y + 54y^2x\}\]

Thus, \[(2x + 3y)^3 = 8x^3 + 36x^2y + 54y^2x + 27y^3\]

Example 13

Expand \[(5 - 6x)^3\]

Following the pattern of the cube of the difference of two terms,

\[(5 - 6x)^3 = 5^3 - (6x)^3 - \{3 \times 5 \times 6x (5 - 6x)\}\]

\[= 125 - 216x^3 - \{90x (5 - 6x)\}\]

\[= 125 - 216x^3 - \{450x - 540x^2\}\]

Thus, \[(5 - 6x)^3 = 125 - 450x + 540x^2 - 216x^3\]

Example 14

Expand \[(x + 2y + 3z)^2\]

Following the pattern of the square of three terms,

\[(x + 2y + 3z)^2 = x^2 + (2y)^2 + (3z)^2 + 2(2xy + 3xz + 6yz)\]

\[= x^2 + 4y^2 + 9z^2 + 4xy + 6xz + 12yz\]

Thus, \[(x + 2y + 3z)^2 = x^2 + 4y^2 + 9z^2 + 4xy + 6xz + 12yz\]

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