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.
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) = 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.
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7th Grade Math Problems
8th Grade Math Practice
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