Question

Expand each of the following, using suitable identities:
$(i) {\left( {x + 2y + 4z} \right)^2}$ $(ii) {\left( {2x - y + z} \right)^2}$ $(iii) {\left( { - 2x + 3y + 2z} \right)^2}$ $(iv) {\left( {3a - 7b - c} \right)^2}$ $(v) {\left( { - 2x + 5y - 3z} \right)^2}$ $(iv) {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2}$

Answer 1

Hint: Let’s substitute the values of a, b and c in the formula given below for determining the square of sum of three numbers and reach the answer by simplifying.
$ \Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$

Complete step-by-step answer:
All the above parts of the question contain the form of ${\left( {a + b + c} \right)^2}$ , and as we know that:
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right),$ we will use this formula for expanding every expression:
$(i)$ If we compare ${\left( {x + 2y + 4z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = x,b = 2y$ and $c = 4z$. Thus, using the formula, we’ll get:
$
   \Rightarrow {\left( {x + 2y + 4z} \right)^2} = {x^2} + {\left( {2y} \right)^2} + {\left( {4z} \right)^2} + 2\left( {x.2y + 2y.4z + x.4z} \right), \\
   \Rightarrow {\left( {x + 2y + 4z} \right)^2} = {x^2} + 4{y^2} + 16{z^2} + 4xy + 16yz + 8xz \\
$
$(ii)$ If we compare ${\left( {2x - y + z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = 2x,b = - y$ and $c = z$. Thus, using the formula, we’ll get:
$
   \Rightarrow {\left( {2x - y + z} \right)^2} = {\left( {2x} \right)^2} + {\left( { - y} \right)^2} + {z^2} + 2\left[ {2x.\left( { - y} \right) + \left( { - y} \right).z + 2x.z} \right], \\
   \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + {y^2} + {z^2} - 4xy - 2yz + 4xz \\
$
$(iii)$ If we compare ${\left( { - 2x + 3y + 2z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = - 2x,b = 3y$ and $c = 2z$. Thus, using the formula, we’ll get:
$
   \Rightarrow {\left( { - 2x + 3y + 2z} \right)^2} = {\left( { - 2x} \right)^2} + {\left( {3y} \right)^2} + {\left( {2z} \right)^2} + 2\left[ {\left( { - 2x} \right).3y + 3y.2z + \left( { - 2x} \right).2z} \right], \\
   \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + 9{y^2} + 4{z^2} - 12xy + 12yz - 8xz \\
$
$(iv)$ If we compare ${\left( {3a - 7b - c} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, in place of a, b and c here we have 3a, -7b and –c respectively. Thus, using the formula, we’ll get:
\[ \Rightarrow {\left( {3a - 7b - c} \right)^2} = {\left( {3a} \right)^2} + {\left( { - 7b} \right)^2} + {\left( { - c} \right)^2} + 2\left[ {3a.\left( { - 7b} \right) + \left( { - 7b} \right).\left( { - c} \right) + 3a.\left( { - c} \right)} \right],\]
\[ \Rightarrow {\left( {3a - 7b - c} \right)^2} = 9{a^2} + 49{b^2} + {c^2} - 42ab + 14bc - 6ac\]
$(v)$ If we compare ${\left( { - 2x + 5y - 3z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = - 2x,b = 5y$ and $c = - 3z$. Thus, using the formula, we’ll get:
\[
   \Rightarrow {\left( { - 2x + 5y - 3z} \right)^2} = {\left( { - 2x} \right)^2} + {\left( {5y} \right)^2} + {\left( { - 3z} \right)^2} + 2\left[ {\left( { - 2x} \right).5y + 5y.\left( { - 3z} \right) + \left( { - 2x} \right).\left( { - 3z} \right)} \right], \\
   \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + 25{y^2} + 9{z^2} - 20xy - 30yz + 12xz \\
\]
$(vi)$ If we compare ${\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, in place of a, b and c here we have $\dfrac{1}{4}a, - \dfrac{1}{2}b$ and 1 respectively. Thus, using the formula, we’ll get:
\[
   \Rightarrow {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2} = {\left( {\dfrac{1}{4}a} \right)^2} + {\left( { - \dfrac{1}{2}b} \right)^2} + {1^2} + 2\left[ {\left( {\dfrac{1}{4}a} \right).\left( { - \dfrac{1}{2}b} \right) + \left( { - \dfrac{1}{2}b} \right).\left( 1 \right) + \left( {\dfrac{1}{4}a} \right).\left( 1 \right)} \right], \\
   \Rightarrow {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2} = \dfrac{1}{{16}}{a^2} + \dfrac{1}{4}{b^2} + 1 - \dfrac{1}{4}ab - b + \dfrac{1}{2}a. \\
\]
Note: If we miss the formula for ${\left( {a + b + c} \right)^2}$, we can apply general multiplication method for expanding the above expressions:
$
   \Rightarrow {\left( {a + b + c} \right)^2} = \left( {a + b + c} \right).\left( {a + b + c} \right), \\
   \Rightarrow {\left( {a + b + c} \right)^2} = a\left( {a + b + c} \right) + b\left( {a + b + c} \right) + c\left( {a + b + c} \right) \\
$
On expansion, we’ll get the same result.