Question

How do you simplify ${{2}^{2}}+\left( 13-\dfrac{8}{4} \right)\centerdot 2-{{\left( 6b-4 \right)}^{2}}$ ?

Answer 1

Hint: In the given question, we are asked to simplify the given term which includes within itself power, products of various terms, square and whole square. In order to simplify this, we would require bodmas rule through which we would be able to apply all the operations easily and will get the correct answer of simplification.

Complete step-by-step solution:
According to the given question we are asked to simplify the given term ${{2}^{2}}+\left( 13-\dfrac{8}{4} \right)\centerdot 2-{{\left( 6b-4 \right)}^{2}}$which can be simplified using bodmas rule which means bracket of division, multiplication, addition and subtraction. Also, we can open the whole square of the last squared term first.
So, after opening the whole square term we will get ${{\left( 6b-4 \right)}^{2}}=36{{b}^{2}}+16-48b$ .
Now, replacing it in the given question we get ${{2}^{2}}+\left( 13-\dfrac{8}{4} \right)\centerdot 2-\left( 36{{b}^{2}}+16-48b \right)$. Similarly, we now need to solve the first bracket as $\left( 13-\dfrac{8}{4} \right)=\dfrac{52-8}{4}=11$ .
Therefore, we can replace the above term also by 11 and now the new term attained would be ${{2}^{2}}+11\centerdot 2-\left( 36{{b}^{2}}+16-48b \right)$. Now, opening the first term and multiplying second term by 2 as given we get $4+22-\left( 36{{b}^{2}}+16-48b \right)$.
Now, what we need to do is that we will remove the last bracket of the given question and bring all terms out of bracket with opposite sign as we have minus sign before the bracket and we will get $26-36{{b}^{2}}-16+48b$, now further simplifying this we get $10-36{{b}^{2}}+48b$.
Therefore, the simplified form of the given equation ${{2}^{2}}+\left( 13-\dfrac{8}{4} \right)\centerdot 2-{{\left( 6b-4 \right)}^{2}}$ is $10-36{{b}^{2}}+48b$.

Note: In the given question we need to remember that we need to simplify the given term one by one and not in one go in order to avoid errors and also make use of the bodmas rule as and when the bracket is present in the term we need to simplify.