Question

Expand the following: \[{\left( {4a - 2b - 3c} \right)^2}\]

Answer 1

Hint: We will solve this question by using the formula of
\[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\], where \[x\], \[y\] and \[z\] are variables.

Complete step-by-step solution:
Step 1: By using the formula of \[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\], we can expand the given expression as below:
\[ \Rightarrow {\left( {4a - 2b - 3c} \right)^2} = {\left( {4a} \right)^2} + {\left( { - 2b} \right)^2} + {\left( { - 3c} \right)^2} + \left( {2 \times \left( {4a} \right) \times \left( { - 2b} \right)} \right) + \left( {2 \times \left( { - 2b} \right) \times \left( { - 3c} \right)} \right) + \left( {2 \times \left( { - 3c} \right) \times \left( {4a} \right)} \right)\]
Where \[x = 4a\], \[y = \left( { - 2b} \right)\] and \[z = \left( { - 3c} \right)\].
Step 2: By opening the brackets and multiplying the terms we get:
\[ \Rightarrow {\left( {4a - 2b - 3c} \right)^2} = {\left( {4a} \right)^2} + {\left( { - 2b} \right)^2} + {\left( { - 3c} \right)^2} - 16ab + 12bc - 24ca\] ……………………. (1)
By replacing the terms as
\[{\left( {4a} \right)^2} = 16{a^2}\], \[{\left( { - 2b} \right)^2} = 4{b^2}\] and \[{\left( { - 3c} \right)^2} = 9{c^2}\] in the above expression (1), we get:
\[ \Rightarrow {\left( {4a - 2b - 3c} \right)^2} = 16{a^2} + 4{b^2} + 9{c^2} - 16ab + 12bc - 24ca\]

\[\because \] The answer is \[{\left( {4a - 2b - 3c} \right)^2} = 16{a^2} + 4{b^2} + 9{c^2} - 16ab + 12bc - 24ca\].

Note: Students should remember some basic formulas for solving these types of questions. Some of them are mentioned below for better understanding:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
\[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\]
The proof of the formula \[{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx\] which we have also used in the above solution is given below:
By writing the term \[{\left( {x + y + z} \right)^2}\] as given below:
\[{\left( {x + y + z} \right)^2} = {\left( {\left( {x + y} \right) + z} \right)^2}\]
By using the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]in the above expression, we get:
\[ \Rightarrow {\left( {x + y + z} \right)^2} = {\left( {x + y} \right)^2} + {z^2} + 2 \times \left( {x + y} \right) \times z\] , where \[a = \left( {x + y} \right)\] and \[b = z\].
Now by again using the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in the above expression, we get:
\[ \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + 2xy + {z^2} + 2 \times \left( {x + y} \right) \times z\]
Simplifying the above expression by doing multiplication inside the brackets we get:
\[ \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + 2xy + {z^2} + 2xz + 2yz\]
We can write the above expression as:
\[ \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2xz\]
Hence proved.