Question

How do you evaluate ${{56}^{2}}$ using special products?
 (a) Using algebraic properties
(b) Using linear formulas
(c) Using trigonometric identities
(d) None of these

Answer 1

Hint: Here we have the value where we need to evaluate ${{56}^{2}}$ using special products. So, we are to use an algebraic formula to get along with our problem. So, in this problem we will use the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. To use that formula, we will use such a quantity that will make our job easier.
So, we will divide 56 into 50 + 6 where the addition and multiplication process become easier. Then by further simplification we will get our result.

Complete step by step solution:
According to the problem, we are to evaluate ${{56}^{2}}$ using special products.
So, to start with, 56 can be written as,$56=\left( 50+6 \right)$.
And we also know from algebraic quantities, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Putting a = 50 and b = 6, we can get,
${{\left( a+b \right)}^{2}}={{\left( 50+6 \right)}^{2}}={{\left( 56 \right)}^{2}}$
Thus, we can conclude that the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$can be used while finding the value of ${{56}^{2}}$.
So, now, ${{56}^{2}}$can be written as, ${{\left( 50+6 \right)}^{2}}$, which is equal to,
 $\Rightarrow {{\left( 50+6 \right)}^{2}}={{50}^{2}}+2\times 50\times 6+{{6}^{2}}$
By further simplification we get,
2500 + 600 +36
So, the final value of ${{56}^{2}}$ is 3136, adding three of them.

So, the correct answer is “Option (a)”.

Note: Using special products in case of certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster.