SQUARE OF A TRINOMIAL 

About "Square of a trinomial"

Sometimes, finding square of a trinomial is a difficult task for some students.

But square of a trinomial is not difficult and it can be determined by using the formula given below.  

( a + b + c )²  = a² + b² + c² + 2ab + 2bc + 2ac  

How can we get this expansion? 

It is very simple. We all know the identity 

( x + y )² = x² + y² + 2xy 

Based on this algebraic identity, we can get the expansion of (a+b+c)² as explained below. 

Let x = a+b. Then, we have 

( a + b + c )²  =  ( x + c )²

( a + b + c )²  =  x² + c² + 2xc

Now, plug x = a + b

( a + b + c )²  =  ( a + b )² + c² +2(a+b)c

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

( a + b + c )²  =  a² + b² + 2ab + c² + 2ac + 2bc 

( a + b + c )²  =  a² + b² + c² 2ab + 2ac + 2bc 

or ( a + b + c )²  =  a² + b² c² 2(ab + ac + bc) 

Now, let us get the expansion of ( a + b + c )², if one or more terms is negative. 

( a + b - c )²  =  [ a + b + (- c )]²

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

=  a² + b² + c² + 2ab - 2ac - 2bc 

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

( a - b + c )²  =  [ a + (-b) + c ]²

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

=  a² + b² + c² - 2ab + 2ac - 2bc 

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

( a - b - c )²  =  [ a + (-b) + (-c) ]²

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

=  a² + b² + c² - 2ab - 2ac + 2bc 

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

Worked out examples

To have better understanding on "Square of a trinomial", let us look at some worked out examples.  

Example 1 :

Expand the following 

( 2x + 3y + 5z )² 

Solution : 

Already, we know ( a + b + c )²  = a² + b² + c² + 2ab + 2bc + 2ac  

If we compare ( 2x + 3y + 5z )² and ( a + b + c )², we have 

a = 2x,  b = 3y,  c = 5z

Let us plug a = 2x,  b = 3y,  c = 5z in the expansion of ( a + b + c )²

So, (2x+3y+5z)² = (2x)² + (3y)² + (5z)² + 2(2x)(3y) + 2(3y(4z) + 2(2x)(5z)

  =  4x² + 9y² + 25z² + 12xy + 24yz + 20xz

Hence, ( 2x + 3y + 5z )² = 4x² + 9y² + 25z² + 12xy + 24yz + 20xz

Let us look at the next example on "Square of a trinomial"

Example 2 :

Expand the following 

( 2x + 3y + 5z )² 

Solution : 

Already, we know ( a + b + c )²  = a² + b² + c² + 2ab + 2bc + 2ac  

If we compare ( 2x + 3y + 5z )² and ( a + b + c )², we have 

a = 2x,  b = 3y,  c = 5z

Let us plug a = 2x,  b = 3y,  c = 5z in the expansion of ( a + b + c )²

So, 

(2x + 3y + 5z)² = (2x)² + (3y)² + (5z)² + 2(2x)(3y) + 2(3y)(4z) + 2(2x)(5z)

  =  4x² + 9y² + 25z² + 12xy + 24yz + 20xz

Hence, ( 2x + 3y + 5z )² = 4x² + 9y² + 25z² + 12xy + 24yz + 20xz

Let us look at the next example on "Square of a trinomial"

Example 3 :

Expand the following 

( 2l + 3m  - 4n )² 

Solution : 

Already, we know ( a + b - c )²  = a² + b² + c² + 2ab - 2bc - 2ac  

If we compare ( 2l + 3m - 4n )² and ( a + b - c )², we have 

a = 2l,  b = 3m,  c = 4n

Let us plug a = 2l,  b = 3m,  c = 4n in the expansion of ( a + b - c )²

So,

(2l+ 3m - 4n)² = (2l)² + (3m)² + (4n)² + 2(2l)(3m) - 2(3m)(4n) - 2(2l)(4n)

  =  4l² + 9m² + 16l² + 12lm - 24mn - 16ln

Hence, (2l+ 3m - 4n)²   =  4l² + 9m² + 16l² + 12lm - 24mn - 16ln

Let us look at the next example on "Square of a trinomial"

Example 4 :

Expand the following 

( x - 2y + 3z )² 

Solution : 

Already, we know ( a - b + c )²  = a² + b² + c² - 2ab - 2bc + 2ac  

If we compare ( x - 2y + 3z )² and ( a + b - c )², we have 

a = x,  b = 2y,  c = 3z

Let us plug a = x,  b = 2y,  c = 3z in the expansion of ( a + b - c )²

So,

(x -2y + 3z)² = x² + (2y)² + (3z)² - 2(x)(2y) - 2(2y)(3z) + 2(x)(3z)

  =  x² + 4y² + 9z² - 4xy - 12yz + 6xz

Hence, ( x - 2y + 3z )²   =  x² + 4y² + 9z² - 4xy - 12yz + 6xz

Let us look at the next example on "Square of a trinomial"

Example 5 :

Expand the following 

( 3p - 2q - 5r )² 

Solution : 

Already, we know ( a - b - c )²  = a² + b² + c² - 2ab + 2bc - 2ac  

If we compare ( 3p - 2q - 5r )² and ( a - b - c )², we have 

a = 3p,  b = 2q,  c = 5r

Let us plug a = 3p,  b = 2q,  c = 5r in the expansion of ( a - b - c )²

So,

(3p -2q - 5r)² = (3p)² + (2q)² + (5r)² - 2(3p)(2q) + 2(2q)(5r) - 2(3p)(5r)

  =  9p² + 4q² + 25r² - 12pq + 20qr - 30pr

Hence, (3p -2q - 5r)²   =  9p² + 4q² + 25r² - 12pq + 20qr - 30pr

Let us look at the next example on "Square of a trinomial"

Example 6 :

If a +b + c = 25 and  ab + bc + ca = 25, find the value of a² + b² + c². 

Solution : 

Already, we know ( a + b + c )²  = a² + b² + c² + 2(ab + bc + ac)  

Let us plug  (a +b + c) = 25 and  (ab + bc + ca) = 25

(25)²  = a² + b² + c² + 2(59)

625  = a² + b² + c² + 118

507 = a² + b² + c²

Hence, a² + b² + c² = 507

Let us look at the next example on "Square of a trinomial"

Example 7 :

If a + b + c = 36 and  a² + b² + c² = 676, then  find the value of 

ab + bc + ca. 

Solution : 

Already, we know ( a + b + c )²  = a² + b² + c² + 2(ab + bc + ac)  

Let us plug  (a +b + c) = 36 and  (a² + b² + c²) = 676

(36)²  = 676 + 2(ab + bc + ac)

1296  = 676 + 2(ab + bc + ac)

620 = 2(ab + bc + ac)

310 = ab + bc + ac

Hence, ab + bc + ac = 310

Shortcut to remember the expansion of square of a trinomial with negative sign

We can remember the expansion of the identities like (a+b)² (a+b+c)², (a+b+c)³. In square of a trinomial, if one or more terms is negative, how can we remember the expansion?

This question has been answered in the following two cases. 

Case 1 :

For example, let us consider the identity of (a + b + c)²

We can easily remember the expansion of (a + b + c)². 

If "c" is negative, we will have (a + b - c)²

How can we remember the expansion of (a + b - c)² ?

It is very simple. 

In the terms of the expansion, a², b², c²,  ab, bc, ca, let us consider the terms in which we find "c"

They are c², bc, ca .

Even if we take negative sign for "c", the sign of  will be positive. Because it has even power "2"

The terms bc, ca will be negative, Because both "b" and "a" are multiplied by "c" which is negative.  

Finally, we have 

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

Case 2 :

In (a+b+c)², if both "b" and "c" are negative, we will have (a - b - c)²

How can we remember the expansion of (a - b - c)² ?

It is very simple. 

In the terms of the expansion, a², b², c²,  ab, bc, ca, let us consider the terms in which we find "b" and "c"

They are b², c², ab,  bc, ca.

Even if we take negative sign for "b" and "c", the sign of b² and  will be positive. Because they even power "2". 

The terms "ab" and "ca" will be negative.

Because, in "ab""a" is multiplied by "b" which is negative.  

Because, in "ca""a" is multiplied by "c" which is negative.  

The term "bc" will be positive.

Because, in "bc", both "b" and "c" are negative.                                           (negative x negative = positive)  

Finally, we have 

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

After having gone through the stuff and examples explained, we hope that the students would have understood,  how to expand square of a trinomial using the identity.  

Apart from the stuff and examples, if you want to know more about "Square of a trinomial", please click here. 

Related topics :

Algebraic identities

Algebra formula calculator

HTML Comment Box is loading comments...