Question

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

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:
$$\begin{gathered} \Rightarrow {\left(x+2y+4z\right)}^2=x^2+{\left(2y\right)}^2+{\left(4z\right)}^2+2\left(x.2y+2y.4z+x.4z\right), \\ \Rightarrow {\left(x+2y+4z\right)}^2=x^2+4y^2+16z^2+4xy+16yz+8xz \end{gathered}$$
$(ii)$ 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:
$$\begin{gathered} \Rightarrow {\left(2x-y+z\right)}^2={\left(2x\right)}^2+{\left(-y\right)}^2+z^2+2\left[2x.\left(-y\right)+\left(-y\right).z+2x.z\right], \\ \Rightarrow {\left(x+2y+4z\right)}^2=4x^2+y^2+z^2-4xy-2yz+4xz \end{gathered}$$
$(iii)$ 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:
$$\begin{gathered} \Rightarrow {\left(-2x+3y+2z\right)}^2={\left(-2x\right)}^2+{\left(3y\right)}^2+{\left(2z\right)}^2+2\left[\left(-2x\right).3y+3y.2z+\left(-2x\right).2z\right], \\ \Rightarrow {\left(x+2y+4z\right)}^2=4x^2+9y^2+4z^2-12xy+12yz-8xz \end{gathered}$$
$(iv)$ 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:
$$\Rightarrow {\left(3a-7b-c\right)}^2={\left(3a\right)}^2+{\left(-7b\right)}^2+{\left(-c\right)}^2+2\left[3a.\left(-7b\right)+\left(-7b\right).\left(-c\right)+3a.\left(-c\right)\right],$$
$$\Rightarrow {\left(3a-7b-c\right)}^2=9a^2+49b^2+c^2-42ab+14bc-6ac$$
$(v)$ 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:
$$\begin{gathered} \Rightarrow {\left(-2x+5y-3z\right)}^2={\left(-2x\right)}^2+{\left(5y\right)}^2+{\left(-3z\right)}^2+2\left[\left(-2x\right).5y+5y.\left(-3z\right)+\left(-2x\right).\left(-3z\right)\right], \\ \Rightarrow {\left(x+2y+4z\right)}^2=4x^2+25y^2+9z^2-20xy-30yz+12xz \end{gathered}$$
$(vi)$ If we compare ${\left({\displaystyle \frac{1}{4}}a-{\displaystyle \frac{1}{2}}b+1\right)}^2$ with ${\left(a+b+c\right)}^2$, in place of a, b and c here we have ${\displaystyle \frac{1}{4}}a,-{\displaystyle \frac{1}{2}}b$ and 1 respectively. Thus, using the formula, we’ll get:
$$\begin{gathered} \Rightarrow {\left({\displaystyle \frac{1}{4}}a-{\displaystyle \frac{1}{2}}b+1\right)}^2={\left({\displaystyle \frac{1}{4}}a\right)}^2+{\left(-{\displaystyle \frac{1}{2}}b\right)}^2+1^2+2\left[\left({\displaystyle \frac{1}{4}}a\right).\left(-{\displaystyle \frac{1}{2}}b\right)+\left(-{\displaystyle \frac{1}{2}}b\right).\left(1\right)+\left({\displaystyle \frac{1}{4}}a\right).\left(1\right)\right], \\ \Rightarrow {\left({\displaystyle \frac{1}{4}}a-{\displaystyle \frac{1}{2}}b+1\right)}^2={\displaystyle \frac{1}{16}}{a}^2+{\displaystyle \frac{1}{4}}{b}^2+1-{\displaystyle \frac{1}{4}}ab-b+{\displaystyle \frac{1}{2}}a. \end{gathered}$$
Note: If we miss the formula for ${\left(a+b+c\right)}^2$, we can apply general multiplication method for expanding the above expressions:
$$\begin{gathered} \Rightarrow {\left(a+b+c\right)}^2=\left(a+b+c\right).\left(a+b+c\right), \\ \Rightarrow {\left(a+b+c\right)}^2=a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right) \end{gathered}$$
On expansion, we’ll get the same result.