Question

1)Select a suitable identity and find the following products:

1)(7d-9e)(7d-9e)
2)(3t+9s)(3t+9s)
3)(kl-mn)(kl-mn)
4)(6x+5)(6x+6)
5)(2b-a)(2b+c)
​

Answer 1

refer to above attachments for solution↑

Answer 2

(i) (7d - 9e)(7d - 9e)

For this question we can consider (7d - 9e)(7d - 9e) as (7d - 9e)²

Now we shall use this identity for this question ⇒ (a - b)² = a² - 2ab + b²

Here a = 7d and b = 9e

(7d - 9e)²

(7d)² - (2 × 7d × 9e) + (9e)²

Now we will square the constants and let the power of two remain on the variables since we don't know the value of the variables.

(7d)² - (2 × 7d × 9e) + (9e)²

49d² - (2 × 7d × 9e) + 81e²

Now we will find the value in the brackets.

49d² - (2 × 7d × 9e) + 81e²

49d² - (2 × 63de) + 81e²

49d² - 126de + 81e²

We can't further simplify.

∴ (7d - 9e)(7d - 9e) = 49d² - 126de + 81e²

$\rule{300}{1}$

(ii) (3t + 9s)(3t + 9s)

For this question we can consider (3t + 9s)(3t + 9s) as (3t + 9s)²

Now we shall use this identity for this question ⇒ (a + b)² = a² + 2ab + b²

Here a = 3t and b = 9s

(3t + 9s)²

(3t)² + (2 × 3t × 9s) + (9s)²

Now we will square the constants and let the power of two remain on the variables since we don't know the value of the variables.

(3t)² + (2 × 3t × 9s) + (9s)²

9t² + (2 × 3t × 9s) + 81s²

Now we will find the value in the brackets.

9t² + (2 × 3t × 9s) + 81s²

9t² + (2 × 27st) + 81s²

9t² + 54st + 81s²

We can't further simplify.

∴ (3t + 9s)(3t + 9s) = 9t² + 54st + 81s²

$\rule{300}{1}$

(iii) (kl - mn)(kl - mn)

For this question we can consider (kl - mn)(kl - mn) as (kl - mn)²

Now we shall use this identity for this question ⇒ (a - b)² = a² - 2ab + b²

Here a = kl and b = mn

(kl - mn)²

(kl)² - (2 × kl × mn) + (mn)²

Now we will simplify the variables by distributing the variables.

(kl)² - (2 × kl × mn) + (mn)²

k²l² - (2 × klmn) + m²n²

Now we will find the value of the brackets.

k²l² - (2 × klmn) + m²n²

k²l² - 2klmn + m²n²

We can't further simplify.

∴ (kl - mn)(kl - mn) = k²l² - 2klmn + m²n²

$\rule{300}{1}$

(iv) (6x + 5)(6x + 6)

Over here we are multiplying a binomial by a binomial. So let's solve.

(6x + 5)(6x + 6)

Now we will simplify it in an easier form to solve.

(6x + 5)(6x + 6)

6x (6x + 6) + 5 (6x + 6)

6x (6x) + 6x (6) + 5 (6x) + 5 (6)

We will distribute and combine like terms and multiply.

6x (6x) + 6x (6) + 5 (6x) + 5 (6)

(6)(6)(x)(x) + (6)(6)(x) + (5)(6)(x) + (5)(6)

36x² + 36x + 30x + 30

Since we see like terms we will simplify more.

36x² + 36x + 30x + 30

36x² + 66x + 30

We can't further simplify.

∴ (6x+5)(6x+6) = 36x² + 66x + 30

$\rule{300}{1}$

(v) (2b - a)(2b + c)

Over here we are multiplying a binomial by a binomial. So let's solve.

(2b - a)(2b + c)

Now we will simplify it in an easier form to solve.

(2b - a)(2b + c)

2b (2b + c) - a (2b + c)

2b (2b) + 2b (c) - a (2b) - a (c)

We will distribute and combine like terms and multiply.

2b (2b) + 2b (c) - a (2b) - a (c)

(2)(2)(b)(b) + (2)(b)(c) - (2)(a)(b) - (a)(c)

4b² + 2bc - 2ab - ac

We can't further simplify.

∴ (2b-a)(2b+c) = 4b² + 2bc - 2ab - ac

$\rule{300}{1}$