**Expansion of Trinomial with Power 2 :**

In this section, you will learn how to expand a trinomial with power 2.

(a + b + c)^{2} = (a + b + c)(a + b + c)

(a + b + c)^{2} = a^{2} + ab + ac + ab + b^{2 }+ bc + ac + bc + c^{2}

**(a + b + c) ^{2} = a^{2} + b^{2 }+ c^{2 }+ 2ab + 2bc + 2ac**

To get expansion for (a + b - c)^{2}, let us consider the expansion of (a + b + c)^{2}.

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac**

In (a + b + c)^{2}, if c is negative, then we have

(a + b - c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find 'c'.

They are c^{2}, bc, ca.

Even if we take negative sign for 'c' in c^{2}, the sign of c^{2} will be positive. Because it has even power 2.

The terms bc, ac will be negative. Because both 'b' and 'a' are multiplied by 'c' that is negative.

Finally, we have

**(a + b - c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab - 2bc - 2ac**

To get expansion for (a - b + c)^{2}, let us consider the expansion of (a + b + c)^{2}.

The expansion of (a + b + c)^{2 }is

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca**

In (a + b + c)^{2}, if b is negative, then we have

(a - b + c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find "b".

They are b^{2}, ab, bc.

Even if we take negative sign for 'b' in b^{2}, the sign of b^{2} will be positive. Because it has even power 2.

The terms ab, bc will be negative. Because both 'a' and 'c' are multiplied by 'b' that is negative.

Finally, we have

**(a - b + c) ^{2} = a^{2} + b^{2} + c^{2} - 2ab - 2bc + 2ac**

To get the expansion of (a - b - c)^{2}, let us consider the expansion of (a + b + c)^{2}.

The expansion of (a + b + c)^{2 }is

**(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca**

In (a + b + c)^{2}, if b and c are negative, then we have

(a - b - c)^{2}

In the terms of the expansion for (a + b + c)^{2}, consider the terms in which we find 'b' and 'c'.

They are b^{2}, c^{2}, ab, bc, ac.

Even if we take negative sign for 'b' in b^{2} and negative sign for 'c' in c^{2}, the sign of both b^{2 }and c^{2} will be positive. Because they have even power 2.

The terms 'ab' and 'ac' will be negative.

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

Because, in 'ac', 'a' is multiplied by "c" that is negative.

The term 'bc' will be positive.

Because, in 'bc', both 'b' and 'c' are negative.

That is,

negative ⋅ negative = positive

Finally, we have

**(a - b - c) ^{2} = a^{2} + b^{2} + c^{2} - 2ab + 2bc - 2ac**

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

(a + b - c)^{2} = a^{2} + b^{2} + c^{2} + 2ab - 2bc - 2ca

(a - b + c)^{2} = a^{2} + b^{2} + c^{2} - 2ab - 2bc + 2ca

(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2ab + 2bc - 2ca

Instead of memorizing all the above formulas, we may memorize the first formula and we may apply values of b and c along with signs.

**Question 1 :**

(i) Expand following :

(2x + 3y + 4z)^{2}

**Solution :**

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

a = 2x, b = 3y and c = 4z

= (2x)^{2}+(3y)^{2}+(4z)^{2}+2(2x)(3y)+2(3y)(4z)+2(4z)(2x)

= 4x^{2} + 9y^{2} + 16z^{2} + 12xy + 24yz + 16zx

(ii) Expand the following :

(−p +2q + 3r)^{2}

**Solution : **

a = -p, b = 2q and c = 3r

= (-p)^{2}+(2q)^{2}+(3r)^{2}+2(-p)(2q)+2(2q)(3r)+2(3r)(-p)

= p^{2} + 4q^{2} + 9r^{2} - 4pq + 12qr - 6rp

(x+a) (x+b) (x+c)

= x^{3} + (a + b + c) x^{2} + (ab + bc + ca)x + abc

(iii) Expand following :

(2p + 3)(2p −4)(2p −5)

**Solution :**

x = 2p, a = 3, b = -4 and c = -5

a + b + c = 3 + (-4) + (-5) = -6

ab + bc + ca = 3(-4) + (-4)(-5) + (-5)(3)

= -12 + 20 - 15

= -27 + 20

= -7

abc = 3(-4)(-5) = 60

= x^{3} + (a + b + c) x^{2} + (ab + bc + ca)x + abc

= (2p)^{3} + (-6) (2p)^{2} + (-7)(2p) + 60

= 8p^{3} -24p^{2} -14p + 60

(iv) Expand following :

(3a +1)(3a −2)(3a + 4)

**Solution : **

x = 3a, a = 1, b = -2 and c = 4

a + b + c = 1 + (-2) + 4 = 3

ab + bc + ca = 1(-2) + (-2)(4) + 4(1)

= -2 - 8 + 4

= -10 + 4

= -6

abc = 1(-2)(4) = -8

= x^{3} + (a + b + c) x^{2} + (ab + bc + ca)x + abc

= (3a)^{3} + 3(3a)^{2} + (-6)(3a) - 8

= 27a^{3} + 27a^{2} -18a - 8

After having gone through the stuff given above, we hope that the students would have understood how to expand a trinomial with power 2.

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