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)
= x
^{2} + xb + xa + ab
= x
^{2} + x (b + a) + ab
Therefore, (x + a) (x + b) = x
^{2} + x(a + b) + ab
Similarly,
● (x + a) (x  b) = (x + a) [x + (b)]
= x
^{2} + x [a + (b)] + a × (b)
= x
^{2} + x (a – b) – ab
Therefore, (x + a) (x  b) = x
^{2} + x (a – b) – ab
● (x  a) (x + b) = [x + (a)] (x + b)
= x
^{2} + x (a + b) + (a) (b)
= x
^{2} + x (b – a) – ab
Therefore, (x  a) (x + b) = x
^{2} + x (b – a) – ab
● (x  a) (x  b) = [x + (a)] [x + (b)]
= x
^{2} + x [(a) + (b) + (a) (b)]
= x
^{2} + x (a  b) + ab
= x
^{2} – x (a + b) + ab
Therefore, (x  a) (x  b) = x
^{2} – x (a + b) + ab
Workedout 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) = x
^{2} + x(a + b) + ab
Here, a = 2 and b = 5
= (y)
^{2} + y(2 + 5) + 2 × 5
= y
^{2} + 7y + 10
Therefore (x + 2) (x + 5) = y
^{2} + 7y + 10
(ii) (p – 2) (p – 3)
Solution:
We know, [x + (a)] [x + ( b)] = x
^{2} + x [( a) + ( b)] + (a) (b)
Therefore, (p – 2) (p – 3) = [p + ( 2)] [p + ( 3)]
Here, a = 2 and b = 3
[p + ( 2)] [p + ( 3)]
= p
^{2} + p [(2) + (3)] + (2) (3)
= p
^{2} + p (2  3) + 6
= p
^{2} – 5p + 6
Therefore, (p – 2) (p – 3) = p
^{2} – 5p + 6
(iii) (m + 3) (m – 2)
Solution:
We know, [x + a] [x + (b)] = x
^{2} + x [a + (b)] + a (b)
Therefore, (m + 3) (m – 2) = (m + 3) [m + (2)]
Here, a = 3, b= 2
(m + 3) [m + (2)]
= m
^{2} + m [3 + (2)] + (3) (2)
= m
^{2} + m [3  2] + (6)
= m
^{2} + m (1)  6
= m
^{2} + m – 6
Therefore (m + 3) (m – 2) = m
^{2} + 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)] = x
^{2} + 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) = x
^{2} + 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) = x
^{2} + 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
From Product of Two Binomials whose First Terms are Same and Second Terms are Different to HOME PAGE
Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.
Share this page:
What’s this?

