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)

                     = a² – ab + 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.

7th Grade Math Problems

8th Grade Math Practice

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