Question

How do you multiply ${{\left( x-6 \right)}^{2}}$?

Answer 1

Hint: We are used to multiplying ${{\left( x-6 \right)}^{2}}$. We will start by observing the number of terms that are present in the given expression. We will then try to find the suitable identity to solve this. We will multiply using the basic approach and then we will be using the algebraic identity also to get the different ways possible to solve this problem.

Complete step by step solution:
We are given ${{\left( x-6 \right)}^{2}}$. As we can see that the power is 2, so it means that we have to multiply $x-6$ with itself. There are various ways to solve this problem. We will start by using the basic approach in which we multiply $x-6$ with itself by the multiplication method in which the single term of the first bracket is multiplied by each term of the second bracket. So, we get,
$\left( x-6 \right)\left( x-6 \right)=x\left( x-6 \right)-6\left( x-6 \right)$
Opening the brackets, we get,
$\Rightarrow x\times x+x\left( -6 \right)-6\left( x \right)-6\left( -6 \right)$
Simplifying it we get,
$\Rightarrow {{x}^{2}}-6x-6x+36$
Now, adding the like terms, we get,
$\Rightarrow {{x}^{2}}-12x+36$
So, we get,
$\Rightarrow {{\left( x-6 \right)}^{2}}={{x}^{2}}-12x+36$
Now, let us work another way to solve it. In this we use the binomial identity,
${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$
So, applying ${{\left( A-B \right)}^{2}}$ on ${{\left( x-6 \right)}^{2}}$, we consider $x=A,y=B$, so we get,
$\Rightarrow {{\left( x-6 \right)}^{2}}={{x}^{2}}+2\times x\times 6+{{6}^{2}}$
Simplifying, we get,
$\Rightarrow {{x}^{2}}-12x+36$
So, we get,
${{\left( x-6 \right)}^{2}}={{x}^{2}}-12x+36$

Note: We must remember that when we multiply variables, their powers are added up, that is $x\times x={{x}^{2}}$.So, one should not write it as $x\times x=2x$. Also remember that only like terms can be added. So, adding $4{{x}^{2}}-12y=8{{x}^{2}}y$ is incorrect as $4{{x}^{2}}$ and $12y$ are not like terms at all. Also, we need to be careful that ${{\left( 2x \right)}^{2}}\ne 4x$ as squares are applied over both 2 as well as $x$. We should also remember that when two negative terms are added, the term gets added and the sign $\left( - \right)$ stays, for example, $-6x-6x=-12x$.