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)
= a
^{2} –
ab +
ba + b
^{2}
= a
^{2} – b
^{2}
Therefore (a + b) (a – b) = a
^{2} – b
^{2}
(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.
Workedout 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) = a
^{2} – b
^{2}
Here a = 2x and b= 7y
= (2x)
^{2} – (7y)
^{2}
= 4x
^{2} – 49y
^{2}
Therefore, (2x + 7y)(2x – 7y) = 4x
^{2} – 49y
^{2}
2. Evaluate 50
^{2} – 49
^{2} using the identity
Solution:
We know a
^{2} – b
^{2} = (a + b)(a – b)
Here a = 50, b = 49
= (50 + 49) (50 – 49)
= 99 × 1
= 99
Therefore, 50
^{2} – 49
^{2} = 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) = a
^{2} – b
^{2}
= (60)
^{2} – (3)
^{2}
= 3600 – 9
= 3591
Therefore, 63 × 57 = 3591
4. Find the value of x if 23
^{2} – 17
^{2} = 6x
Solution:
We know a
^{2} – b
^{2} = (a + b) (a – b)
Here a = 23 and b = 17
Therefore 23
^{2} – 17
^{2} = 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) = a
^{2} – b
^{2}
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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