Question

Simplify the expression ${{\left( x+5 \right)}^{2}}$ .

Answer 1

Hint: For these kinds of questions, all we need are simple and basic algebraic formulae. We are going to use the formula of ${{\left( a+b \right)}^{2}}$.We are going to compare the standard formula with the given equation and find the value of $a,b$ here. After using the formula, we will expand it and see what kind of polynomial we are ending up with.

Complete step-by-step solution:
We all are well aware of the formula of ${{\left( a+b \right)}^{2}}$.
It is the following :
$\Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .
Now let us compare the standard formula with the given equation.
Upon comparing, the value of $a$ we get is $x$ and the value of $b$ we get is $5$.
Now let us use the formula and expand to see the kind of polynomial we get.
Upon using and expanding, we get the following :
$\begin{align}
  & \Rightarrow {{\left( x+5 \right)}^{2}}={{x}^{2}}+{{5}^{2}}+2\times 5\times x \\
 & \Rightarrow {{\left( x+5 \right)}^{2}}={{x}^{2}}+25+10x \\
 & \Rightarrow {{\left( x+5 \right)}^{2}}={{x}^{2}}+10x+25 \\
\end{align}$
We are getting a quadratic equation.
$\left( x+5 \right)$ is nothing but the factor of ${{x}^{2}}+10x+25$.
We can also get the same result by using binomial theorem. In fact, the formula of ${{\left( a+b \right)}^{2}}$ is derived from the binomial theorem.
Let us take the expression from the question itself and power it to $n$.
Upon doing so, we get the following :
$\Rightarrow {{\left( x+5 \right)}^{n}}$
We all know how the expansion in binomial theorem goes. Let us use that.
Upon using, we get the following :
\[\Rightarrow {{\left( x+5 \right)}^{n}}={}^{n}{{C}_{0}}\left( {{x}^{n}} \right){{5}^{0}}+{}^{n}{{C}_{1}}\left( 5 \right){{\left( x \right)}^{n-1}}+{}^{n}{{C}_{2}}{{\left( 5 \right)}^{2}}{{\left( x \right)}^{n-2}}+.....+{}^{n}{{C}_{n}}{{\left( 5 \right)}^{n}}{{x}^{0}}\]
We know that the value of $n$in this question is $2$. Let us substitute that.
Upon substituting, we get the following :
\[\begin{align}
  & \Rightarrow {{\left( x+5 \right)}^{n}}={}^{n}{{C}_{0}}\left( {{x}^{n}} \right){{5}^{0}}+{}^{n}{{C}_{1}}\left( 5 \right){{\left( x \right)}^{n-1}}+{}^{n}{{C}_{2}}{{\left( 5 \right)}^{2}}{{\left( x \right)}^{n-2}}+.....+{}^{n}{{C}_{n}}{{\left( 5 \right)}^{n}}{{x}^{0}} \\
 & \Rightarrow {{\left( x+5 \right)}^{2}}={}^{2}{{C}_{0}}\left( {{x}^{2}} \right){{5}^{0}}+{}^{2}{{C}_{1}}\left( 5 \right){{\left( x \right)}^{2-1}}+{}^{2}{{C}_{2}}{{\left( 5 \right)}^{2}}{{\left( x \right)}^{2-2}} \\
 & \Rightarrow {{\left( x+5 \right)}^{2}}=\left( {{x}^{2}} \right)+2\times \left( 5 \right)\left( x \right)+{{\left( 5 \right)}^{2}} \\
 & \Rightarrow {{\left( x+5 \right)}^{2}}={{x}^{2}}+10x+25 \\
\end{align}\]
We got the same result.

Note: For very small values of $n$ such as $2,3,4$ memorizing the formulae will help us save time in the exam. But for greater values of $n$ , binomial theorem is going to come in handy. But we should remember all the values of ${}^{n}{{C}_{r}}$ so as to solve them. We should be careful while solving as there is a scope for a lot of calculation errors.