Product of Two Binomials whose First Terms are Same and Second Terms are Different

How to find the product of two binomials whose first terms are same and second terms are different?



(x + a) (x + b) = x (x + b) + a (x + b)

                    = x2 + xb + xa + ab

                    = x2 + x (b + a) + ab

Therefore, (x + a) (x + b) = x2 + x(a + b) + ab

Similarly,

(x + a) (x - b) = (x + a) [x + (-b)]

                    = x2 + x [a + (-b)] + a × (-b)

                    = x2 + x (a – b) – ab

Therefore, (x + a) (x - b) = x2 + x (a – b) – ab


(x - a) (x + b) = [x + (-a)] (x + b)

                    = x2 + x (-a + b) + (-a) (b)

                    = x2 + x (b – a) – ab

Therefore, (x - a) (x + b) = x2 + x (b – a) – ab


(x - a) (x - b) = [x + (-a)] [x + (-b)]

                    = x2 + x [(-a) + (-b) + (-a) (-b)]

                    = x2 + x (-a - b) + ab

                    = x2 – x (a + b) + ab

Therefore, (x - a) (x - b) = x2 – x (a + b) + ab

Worked-out examples on the product of two binomials whose first terms are same and second terms are different:

1. Find the product of the following using identities:

(i) (y + 2) (y + 5)                               

Solution:

We know, (x + a) (x + b) = x2 + x(a + b) + ab

Here, a = 2 and b = 5

= (y)2 + y(2 + 5) + 2 × 5

= y2 + 7y + 10

Therefore (x + 2) (x + 5) = y2 + 7y + 10


(ii) (p – 2) (p – 3)

Solution:

We know, [x + (-a)] [x + (- b)] = x2 + x [(- a) + (- b)] + (-a) (-b)

Therefore, (p – 2) (p – 3) = [p + (- 2)] [p + (- 3)]

Here, a = -2 and b = -3

[p + (- 2)] [p + (- 3)]

= p2 + p [(-2) + (-3)] + (-2) (-3)

= p2 + p (-2 - 3) + 6

= p2 – 5p + 6

Therefore, (p – 2) (p – 3) = p2 – 5p + 6


(iii) (m + 3) (m – 2)

Solution:

We know, [x + a] [x + (-b)] = x2 + x [a + (-b)] + a (-b)

Therefore, (m + 3) (m – 2) = (m + 3) [m + (-2)]

Here, a = 3, b= -2

(m + 3) [m + (-2)]

= m2 + m [3 + (-2)] + (3) (-2)

= m2 + m [3 - 2] + (-6)

= m2 + m (1) - 6

= m2 + m – 6

Therefore (m + 3) (m – 2) = m2 + m – 6



2. Use the identity (x + a) (x + b) to find the product 63 × 59

Solution:

63 × 59 = (60 + 3) (60 – 1)

= [60 + 3] [60 + ( - 1)]

We know that (x + a) [x + (-b)] = x2 + x [a – (-b)] + (a) (-b)

Here, x = 60, a = 3, b = -1

Therefore, (60 + 3) (60 – 1) = (60)2 + 60 [3 + (-1)] + (3) (-1)

                                      = 3600 + 60 [3 – 1] + (-3)

                                      = 3600 + 60 × 2 - 3

                                      = 3600 + 120 – 3

                                      = 3720 – 3

                                      = 3717

Therefore, 63 × 59 = 3717


3. Evaluate the product without direct multiplication:

(i) 91 × 93           

Solution:

91 × 93 = (90 + 1) (90 + 3)     

We know, (x + a) (x + y) = x2 + x (a + b) + ab}

Here, x = 90, a = 1, b = 3

Therefore, (90 + 1) (90 + 3) = (90)2 + 90 (1 + 3) + 1 × 3

                                       = 8100 + 90 × 4 + 3

                                       = 8100 + 360 + 3

                                       = 8460 + 3

                                       = 8463

Therefore, 91 × 93 = 8463


(ii) 305 × 298

Solution:

305 × 298 = (300 + 5) (300 – 2)       

We know, (x + a) (x - y) = x2 + x (a - b) - ab}

Here, x = 300, a = 5, b = 2

Therefore, (300 + 5) (300 – 2) = (300)2 + 300 [5 + (-2)] + (5)(-2)

                                          = 90000 + 300 × 3 – 10

                                          = 90000 + 900 – 10

                                          = 90900 – 10

                                          = 90890

Therefore, 305 × 298 = 90890


Thus, we learn to use the identity to find the product of two binomials whose first terms are same and second terms are different.

7th Grade Math Problems

8th Grade Math Practice

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