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 = a2 + 2ax + x2

                         = a2 + 2a (b + c) + (b + c)2

                         = a2 + 2ab + 2ac + (b2 + c2 + 2bc)

                         = a2 + b2 + c2 + 2ab + 2bc + 2ca

Therefore, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca




● (a + b - c)2 = [a + b + (-c)]2

                   = a2 + b2 + (-c)2 + 2ab + 2 (b) (-c) + 2 (-c) (a)

                   = a2 + b2 + c2 + 2ab – 2bc - 2ca

Therefore, (a + b - c)2 = a2 + b2 + c2 + 2ab – 2bc - 2ca


● (a - b + c)2 = [a + (- b) + c]2

                   = a2 + (-b2) + c2 + 2 (a) (-b) + 2 (-b) (-c) + 2 (c) (a)

                   = a2 + b2 + c2 – 2ab – 2bc + 2ca

Therefore, (a - b + c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca


● (a - b - c)2 = [a + (-b) + (-c)]2

                   = a2 + (-b2) + (-c2) + 2 (a) (-b) + 2 (-b) (-c) + 2 (-c) (a)

                   = a2 + b2 + c2 – 2ab + 2bc – 2ca

Therefore, (a - b - c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca


Worked-out 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 = = a2 + b2 + c2 + 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)

= 4x2 + 9y2 + 25z2 + 12xy + 30yz + 20zx

Therefore, (2x + 3y + 5z)2 = 4x2 + 9y2 + 25z2 + 12xy + 30yz + 20zx




(ii) (2l – 3m + 4n)2

Solution:

(2l – 3m + 4n)2

We know, (a - b + c)2 = a2 + b2 + c2 – 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)

= 4l2 + 9m2 + 16n2 – 12lm – 24mn + 16nl

Therefore, (2l – 3m + 4n)2 = 4l2 + 9m2 + 16n2 – 12lm – 24mn + 16nl


(iii) (3x – 2y – z)2

Solution:

(3x – 2y – z)2

We know, (a - b - c) 2 = a2 + b2 + c2 – 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)

= 9x2 + 4y2 + z2 – 12xy + 4yz – 6zx



2. Simplify a + b + c = 25 and ab + bc + ca = 59.
    Find the value of a2 + b2 + c2.

Solution:

According to the question, a + b + c = 25

Squaring both the sides, we get

(a+ b + c)2 = (25)2

a2 + b2 + c2 + 2ab + 2bc + 2ca = 625

a2 + b2 + c2 + 2(ab + bc + ca) = 625

a2 + b2 + c2 + 2 × 59 = 625 [Given, ab + bc + ca = 59]

a2 + b2 + c2 + 118 = 625

a2 + b2 + c2 + 118 – 118 = 625 – 118 [subtracting 118 from both the sides]

Therefore, a2 + b2 + c2 = 507

Thus, the formula of square of a trinomial will help us to expand.





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