Square of a Trinomial
How to expand the square of a trinomial?
The square of the sum of three or more
terms can be determined by the formula of the determination of the square of
sum of two terms.
Now we will learn to expand the square of
a trinomial (a + b + c).
Let (b + c) = x
Then (a + b + c)
^{2} = (a + x)
^{2} = a
^{2} + 2ax + x
^{2}
= a
^{2} + 2a (b + c) + (b + c)
^{2}
= a
^{2} + 2ab + 2ac + (b
^{2} + c
^{2} + 2bc)
= a
^{2} + b
^{2} + c
^{2} + 2ab + 2bc + 2ca
Therefore, (a + b + c)
^{2} = a
^{2} + b
^{2} + c
^{2} + 2ab + 2bc + 2ca
● (a + b  c)
^{2} = [a + b + (c)]
^{2}
= a
^{2} + b
^{2} + (c)
^{2} + 2ab + 2 (b) (c) + 2 (c) (a)
= a
^{2} + b
^{2} + c
^{2} + 2ab – 2bc  2ca
Therefore, (a + b  c)
^{2} = a
^{2} + b
^{2} + c
^{2} + 2ab – 2bc  2ca
● (a  b + c)
^{2} = [a + ( b) + c]
^{2}
= a
^{2} + (b
^{2}) + c
^{2} + 2 (a) (b) + 2 (b) (c) + 2 (c) (a)
= a
^{2} + b
^{2} + c
^{2} – 2ab – 2bc + 2ca
Therefore, (a  b + c)
^{2} = a
^{2} + b
^{2} + c
^{2} – 2ab – 2bc + 2ca
● (a  b  c)
^{2} = [a + (b) + (c)]
^{2}
= a
^{2} + (b
^{2}) + (c
^{2}) + 2 (a) (b) + 2 (b) (c) + 2 (c) (a)
= a
^{2} + b
^{2} + c
^{2} – 2ab + 2bc – 2ca
Therefore, (a  b  c)
^{2} = a
^{2} + b
^{2} + c
^{2} – 2ab + 2bc – 2ca
Workedout examples on square of a trinomial:
1. Expand each of the following.
(i) (2x + 3y + 5z)
^{2}
Solution:
(2x + 3y + 5z)
^{2}
We know, (a + b + c)
^{2} = = a
^{2} + b
^{2} + c
^{2} + 2ab + 2bc + 2ca
Here a = 2x, b = 3y and c = 5z
= (2x)
^{2} + (3y)
^{2} + (5z)
^{2} + 2 (2x) (3y) + 2 (3y) (5z) + 2 (5z) (2x)
= 4x
^{2} + 9y
^{2} + 25z
^{2} + 12xy + 30yz + 20zx
Therefore, (2x + 3y + 5z)
^{2} = 4x
^{2} + 9y
^{2} + 25z
^{2} + 12xy + 30yz + 20zx
(ii) (2l – 3m + 4n)
^{2}
Solution:
(2l – 3m + 4n)
^{2}
We know, (a  b + c)
^{2} = a
^{2} + b
^{2} + c
^{2} – 2ab  2bc + 2ca
Here a = 2l, b = 3m and c = 4n
(2l + (3m) + 4n)
^{2}
= (2l)
^{2} + (3m)
^{2} + (4n)
^{2} + 2 (2l) (3m) + 2 (3m) (4n) + 2 (4n) (2l)
= 4l
^{2} + 9m
^{2} + 16n
^{2} – 12lm – 24mn + 16nl
Therefore, (2l – 3m + 4n)
^{2} = 4l
^{2} + 9m
^{2} + 16n
^{2} – 12lm – 24mn + 16nl
(iii) (3x – 2y – z)
^{2}
Solution:
(3x – 2y – z)
^{2}
We know, (a  b  c)
^{2} = a
^{2} + b
^{2} + c
^{2} – 2ab + 2bc – 2ca
Here a = 3x, b = 2y and c = z
[3x + (2y) + (z)]
^{2}
= (3x)
^{2} + (2y)
^{2} + (z)
^{2} + 2 (3x) (2y) + 2 (2y) (z) + 2 (z) (3x)
= 9x
^{2} + 4y
^{2} + z
^{2} – 12xy + 4yz – 6zx
2. Simplify a + b + c = 25 and ab + bc + ca = 59.
Find the value of a
^{2} + b
^{2} + c
^{2}.
Solution:
According to the question, a + b + c = 25
Squaring both the sides, we get
(a+ b + c)
^{2} = (25)
^{2}
a
^{2} + b
^{2} + c
^{2} + 2ab + 2bc + 2ca = 625
a
^{2} + b
^{2} + c
^{2} + 2(ab + bc + ca) = 625
a
^{2} + b
^{2} + c
^{2} + 2 × 59 = 625 [Given, ab + bc + ca = 59]
a
^{2} + b
^{2} + c
^{2} + 118 = 625
a
^{2} + b
^{2} + c
^{2} + 118 – 118 = 625 – 118 [subtracting 118 from both the sides]
Therefore, a
^{2} + b
^{2} + c
^{2} = 507
Thus, the formula of square of a trinomial
will help us to expand.
7th Grade Math Problems
8th Grade Math Practice
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