# Product of Sum and Difference of Two Binomials

How to find the product of sum and difference of two binomials with the same terms and opposite signs?

(a + b) (a – b) = a(a – b) + b (a – b)

= a2ab + ba + b2

= a2 – b2

Therefore (a + b) (a – b) = a2 – b2

(First term + Second term) (First term – Second term) = (First term)2 – (Second term) 2

It is stated as: The product of the binomial sum and difference is equal to the square of the first term minus the square of the second term.

Worked-out examples on the product of sum and difference of two binomials:

1. Find the product (2x + 7y) (2x – 7y) by using the identity.

Solution:

We know (a + b) (a – b) = a2 – b2

Here a = 2x and b= 7y

= (2x)2 – (7y)2

= 4x2 – 49y2

Therefore, (2x + 7y)(2x – 7y) = 4x2 – 49y2

2. Evaluate 502 – 492 using the identity

Solution:

We know a2 – b2 = (a + b)(a – b)

Here a = 50, b = 49

= (50 + 49) (50 – 49)

= 99 × 1

= 99

Therefore, 502 – 492 = 99

3. Simplify 63 × 57 by expressing it as the product of binomial sum and difference.

Solution:

63 × 57 = (60 + 3) (60 – 3)

We know (a + b) (a – b) = a2 – b2

= (60)2 – (3)2

= 3600 – 9

= 3591

Therefore, 63 × 57 = 3591

4. Find the value of x if 232 – 172 = 6x

Solution:

We know a2 – b2 = (a + b) (a – b)

Here a = 23 and b = 17

Therefore 232 – 172 = 6x

(23 + 17)(23 – 17) = 6x

40 × 6 = 6x

240 = 6x

6x/6 = 240/6

Therefore, x = 40

5. Simplify 43 × 37 by expressing it as a difference of two squares.

Solution:

43 × 37 = (40 + 3)( 40 – 3)

We know (a + b) (a – b) = a2 – b2

Here a = 40 and b = 3

= (40)2 – (3)2

= 1600 – 9

= 1591

Therefore, 43 × 37 = 1591

Thus, the product of sum and difference of two binomials is equal to the square of the first term minus the square of the second term.

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