There are different types of numbers that we study in mathematics. In this section, we are going to discuss perfect square numbers. A perfect square is said to be a number that is obtained by multiplying some other number by itself. In algebra, we come across with perfect square expressions or polynomials too. We shall go ahead and learn perfect squares in detail.

Perfect square is an integer which is calculated by squaring another integer or by finding out the product of the other integer with itself. For example, $49$ is a perfect square because it is the square of integers $+7$ or $-7$, that is $(+7)^{2}$ = $49$ and $(-7)^{2}$ = $49$. Also, when we multiply $+7$ with itself or $-7$ with itself, that is $(+7)\ \times\ (+7)$ = $49$ or $(-7)\ \times\ (-7)$ = $49$.

In algebra, polynomial perfect square is when a polynomial is multiplied with itself or it is the square of the polynomial. For example, $3x + 4y$ is a polynomial and the perfect square of this polynomial would be $(3x + 4y)^{2}$ = $9x^{2} + 24xy + 16y^{2}$. It would give the same result when the polynomial $3x + 4y$ is multiplied with itself, $(3x + 4y)\ \times\ (3x + 4y)$ = $9x^{2} + 24xy + 16y^{2}$.

The formula of perfect square is as follows:


a) Square of the integer $p$ is $p^{2}$ or $p\ \times\ p$

b) Perfect Square Trinomial: It s defined as the square of a binomial, where the result we get is a trinomial. The result is in the standard 
form of quadratic equation $ax^{2} + bx + c$. The condition to be a perfect square trinomial is that $b^{2}$ = $4ac$. Following are the perfect square trinomial formulas

i) $(p + q)^{2}$ = $p^{2} + 2pq + q^{2}$, it is the square of a binomial expression which is addition of terms. In this either $p$ or $q$ could be both variables or a variable and a constant.

ii) $(p – q)^{2}$ = $p^{2} – 2pq + q^{2}$, it is the square of a binomial expression which is subtraction of terms. In this either $p$ or $q$ could be both variables or a variable and a constant.
It is a method to factorize a quadratic equation and solve for the unknown variable $x$. Let us study about this method in detail with the help of an example

Example: $2x^{2}\ +\ 20$ = $6x$

Step 1: We rearrange the given equation. This is done by bringing all the $x$ terms on one side and the loose number or the constant on the               other side.

            $2x^{2}\ -\ 6x$ = $-20$
Step 2: The coefficient of the highest power should always be one. If the coefficient of the highest power is not one, then we need to divide                 the whole equation by the coefficient of the highest power.

        $\frac{(2x^{2}\ –\ 6x)}{2}$ = $\frac{-20}{2}$

            $x^{2}\ –\ 3x$ = $-10$
Step 3: Square of the half of the coefficient of the middle term which is the coefficient of the $x$ term is taken and added on both sides of the             equation.

            Coefficient of the middle term or the coefficient of the $x$ term is $-3$. Half of it is $\frac{-3}{2}$.
           
            Squaring, we get $(\frac{-3}{2})^2$ = $\frac{9}{4}$. Adding it to both sides of the equation, we get

            $x^{2}\ –\ 3x\ +$ $\frac{9}{4}$ = $-10\ +$ $\frac{9}{4}$
Step 4: Factorize the perfect square trinomial on the left and simplifying the right, we get

            $X^{2}\ –$ $\frac{2\ \times\ 3}{2\ \times\ x}$ + $\frac{9}{4}$ = $-10\ +$ $\frac{9}{4}$

         $(x – \frac{3}{2})^{2}$ = $\frac{-40}{4}$ + $\frac{9}{4}$

         $(x – \frac{3}{2})^{2}$ = $\frac{-31}{4}$
Step 5: Taking square root on both sides and considering both the + and the – value of the square root.

            $x$ – $\frac{3}{2}$ = $+\ -$ $\sqrt{(\frac{-31}{4})}$
Step 6: Combining like terms and separating out $x$ on the left side, we solve for the unknown variable $x$. The negative radical is                           expressed as an imaginary number

            $x$ = $+\ -$ $\sqrt{\frac{-31i}{2}}$ + $\frac{3}{2}$

            $x$ = $\frac{(3\ +\ i\sqrt{-31})}{2}$, $\frac{(3\ -\ i\sqrt{-31})}{2}$

List of perfect squares is unlimited and would go up to infinity. So, in this article, we would consider the first 50 perfect square numbers and their corresponding factors for convenience.


$1$ = $1\ \ \times\ \ 1$
$4$ = $2\ \ \times\ \ 2$
$9$ = $3\ \times\ 3$
$16$ = $4\ \times\ 4$
$25$ = $5\ \times\ 5$
$36$ = $6\ \times\ 6$
$49$ = $7\ \times\ 7$
$64$ = $8\ \times\ 8$
$81$ = $9\ \times\ 9$
$100$ = $10\ \times\ 10$
$121$ = 1$1\ \times\ 11$
$144$ = $12\ \times\ 12$
$169$ = $13\ \times\ 13$
$196$ = $14\ \times\ 14$
$225$ = $15\ \times\ 15$
$256$ = $16\ \times\ 16$
$289$ = $17\ \times\ 17$
$324$ = $18\ \times\ 18$
$381$ = $19\ \times\ 19$
$400$ = $20\ \times\ 20$
$441$ = $21\ \times\ 21$
$484$ = $22\ \times\ 22$
$529$ = $23\ \times\ 23$
$576$ = $24\ \times\ 24$
$625$ = $25\ \times\ 25$
$676$ = $26\ \times\ 26$
$729$ = $27\ \times\ 27$
$784$ = $28\ \times\ 28$
$841$ = $29\ \times\ 29$
$900$ = $30\ \times\ 30$
$961$ = $31\ \times\ 31$
$1024$ = $32\ \times\ 32$
$1089$ = $33\ \times\ 33$
$1156$ = $34\ \times\ 34$
$1225$ = $35\ \times\ 35$
$1296$ = $36\ \times\ 36$
$1369$ = $37\ \times\ 37$
$1444$ = $38\ \times\ 38$
$1521$ = $39\ \times\ 39$
$1600$ = $40\ \times\ 40$
$1681$ = $41\ \times\ 41$
$1764$ = $42\ \times\ 42$
$1849$ = $43\ \times\ 43$
$1936$ = $44\ \times\ 44$
$2025$ = $45\ \times\ 45$
$2116$ = $46\ \times\ 46$
$2209$ = $47\ \times\ 47$
$2304$ = $48\ \times\ 48$
$2401$ = $49\ \times\ 49$
$2500$ = $50\ \times\ 50$ ...... and the list of perfect square goes on.
Yes, we can say that $25$ is a perfect square because it is an integer which is the square of integer $5$ and $-5$. That is, when we multiply $5$ with itself $5\ \times\ 5$, we get $25$. Also, when we multiply the integer $-5$ with itself $-5\ \times\ -5$, we get $25$.
Example 1: 

Find the perfect square of $x^{2}\ +\ 10x\ +\ 25$.

Solution:

The perfect square applicable is $(a+b)^{2}$ = $a^{2}\ +\ 2ab\ +\ b^{2}$

Here, $a$ = $x$, $2ab$ = $10x$, so $b$ = $5$

So, the perfect square is that of $(a+b)^{2}$ = $(x+5)^{2}$
Example 2:

Find the perfect square of $4x^{2}\ -\ 12x\ +\ 9$.

Solution:

The perfect square applicable is $(a-b)^{2}$ = $a^{2}\ -\ 2ab\ +\ b^{2}$

Here, $a^{2}$ = $4x^{2},\ b^{2}$ = $9$, so $a$ = $2x$ and $b$ = $3$

Therefore, the perfect square is that of $(a-b)^{2}$ = $(2x\ –\ 3)^{2}$
Example 3:

Find the perfect square of $8$.

Solution: 

The perfect square of $8$ is found out by finding out the square of $8$ = $8^{2}$ = $64$ or by multiplying the number $8$ with itself $8\ \times\ 8$ = $64$.
Example 4:

Find the perfect square of $-5$.

Solution: 

The perfect square of $-5$ is found out by finding out the square of $-5$ i.e. $(-5)^{2}$ = $25$ or by multiplying the number $-5$ with itself, $-5\ \times\ -5$ = $25$.