NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 6 — Figure it Out (Pages 142–143)#
To multiply two expressions, multiply every term of the first by every term of the second, then add the like terms:
$$(a + b)(c + d) = ac + ad + bc + bd$$1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid. #
Solution
The middle number $pq$ is in row $p$ and column $q$. The rows above and below are $p - 1$ and $p + 1$, and the columns to the left and right are $q - 1$ and $q + 1$. (For the frame shown, $p = 4$ and $q = 6$.)
| column $q - 1$ | column $q$ | column $q + 1$ | |
|---|---|---|---|
| row $p - 1$ | $(p - 1)(q - 1)$ | $(p - 1)q$ | $(p - 1)(q + 1)$ |
| row $p$ | $p(q - 1)$ | $pq$ | $p(q + 1)$ |
| row $p + 1$ | $(p + 1)(q - 1)$ | $(p + 1)q$ | $(p + 1)(q + 1)$ |
Check with $p = 4$, $q = 6$: the top-left number is $3 \times 5 = 15$ and the bottom-right is $5 \times 7 = 35$, as in the grid.
2. Expand the following products. (i) (3 + u) (v – 3) (ii) 2/3 (15 + 6a) (iii) (10a + b) (10c + d) (iv) (3 – x) (x – 6) (v) (–5a + b) (c + d) (vi) (5 + z) (y + 9) #
(i) (3 + u)(v – 3)
$$= uv + 3v - 3u - 9$$
(ii) 2/3 (15 + 6a)
$$= 10 + 4a$$
(iii) (10a + b)(10c + d)
(iv) (3 – x)(x – 6)
$$= -x^2 + 9x - 18$$
(v) (–5a + b)(c + d)
(vi) (5 + z)(y + 9)
$$= yz + 5y + 9z + 45$$
3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4. #
Solution
Let the numbers be $a$ and $b$. We need:
$$(a + 2)(b - 4) = ab$$$$\Rightarrow ab - 4a + 2b - 8 = ab$$
$$\Rightarrow 2b = 4a + 8$$
$$\Rightarrow b = 2a + 4$$
So any pair with $b = 2a + 4$ works:
| $a$ | $b = 2a + 4$ | $ab$ | $(a + 2)(b - 4)$ |
|---|---|---|---|
| 1 | 6 | $1 \times 6 = 6$ | $3 \times 2 = 6$ |
| 2 | 8 | $2 \times 8 = 16$ | $4 \times 4 = 16$ |
| 3 | 10 | $3 \times 10 = 30$ | $5 \times 6 = 30$ |
∴ Three examples are (1, 6), (2, 8) and (3, 10).
4. Expand (i) (a + ab – 3b²) (4 + b), and (ii) (4y + 7) (y + 11z – 3). #
(i) (a + ab – 3b²)(4 + b)
We multiply each term of the first expression by 4 and then by $b$:
$$(a + ab - 3b^2)(4 + b) = 4a + 4ab - 12b^2 + ab + ab^2 - 3b^3$$$$= 4a + 5ab + ab^2 - 12b^2 - 3b^3$$
(ii) (4y + 7)(y + 11z – 3)
$$= 4y^2 + 44yz - 5y + 77z - 21$$
5. Expand (i) (a – b) (a + b), (ii) (a – b) (a² + ab + b²) and (iii) (a – b)(a³ + a²b + ab² + b³), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding? #
(i) (a – b)(a + b)
$$= a^2 - b^2$$
(ii) (a – b)(a² + ab + b²)
$$= a^3 - b^3$$
(iii) (a – b)(a³ + a²b + ab² + b³)
$$= a^4 - b^4$$
The pattern and the next identity
Pattern: when $a - b$ multiplies a sum of terms whose powers of $a$ go down by one while the powers of $b$ go up, all the middle terms cancel in pairs, leaving only $a^n - b^n$.
The next identity is:
$$(a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5$$Check by expanding:
$$(a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 - a^4b - a^3b^2 - a^2b^3 - ab^4 - b^5$$$$= a^5 - b^5$$