The formula for the difference of squares

Difference of Squares Formula

The difference of squares formula is another key algebraic shortcut that simplifies expressions involving two squared terms subtracted from each other. It is written as:

$(X - Y)^2=X^2 - 2XY + Y^2$
This formula skips the need for full expansion and directly factors the expression. It works for both numerical and algebraic expressions, making it versatile in solving equations and simplifying terms. That is, when we encounter two numbers with a minus sign between them, that is, the difference and they will be in parentheses and raised as a squared expression, we can use this formula.

For $(x - y)^2$ the full expansion would be:
$(x - y)^2=(x-y)(x−y)=x⋅x+x⋅(-y)-y⋅x−y⋅(-y)=x^2-2xy+y^2$

Example:

$(a - 4)^2=$
$a\times a+a\times (-4)+ (-4)\times a + (-4) \times (-4) =$
$a^2+2(-4a)+ (-4)^2 =$
$a^2-8a+16$

Detailed explanation

Start practice

Test yourself on square of difference!

$$ (4b-3)(4b-3) $$

Rewrite the above expression as an exponential summation expression:

Practice more now

Pay attention - The formula also works in non-algebraic expressions or combinations with numbers and unknowns.

Let's look at an example

$(X-7)^2=$
Here we identify two elements between which there is a minus sign and they are enclosed in parentheses and raised to the square as a single expression.
Therefore, we can use the formula for the difference of squares.
We will work according to the formula and pay attention to the minus and plus signs.
We will obtain:
$(X-7)^2=x^2-14x+49$
Indeed, we pronounce the same expression differently using the formula.

If you are interested in this article, you might also be interested in the following articles:

Square

The area of a square

Multiplication of the sum of two elements by the difference between them

The formula for the sum of squares

The formulas that refer to two expressions to the power of 3

In the blog of Tutorela you will find a variety of articles about mathematics.

Examples and exercises with solutions of the difference of squares formula

Exercise #1

Choose the expression that has the same value as the following:

$(x-y)^2$

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

$(x-y)(x-y)=$

$x^2-xy-yx+y^2=$

$x^2-2xy+y^2$

Answer

$x^2-2xy+y^2$

Exercise #2

$x^2+144=24x$

Video Solution

Step-by-Step Solution

Let's solve the given equation:

$x^2+144=24x$

First, let's arrange the equation by moving terms:

$x^2+144=24x \\ x^2-24x+144=0$

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We can do this using the fact that:

$144=12^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation we got in the last step:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2$ indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot12=24x$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}$

Let's summarize the solution of the equation:

$x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}$

Therefore the correct answer is answer C.

Answer

$x=12$

Exercise #3

$x^2=6x-9$

Video Solution

Step-by-Step Solution

Let's solve the given equation:

$x^2=6x-9$

First, let's arrange the equation by moving terms:

$x^2=6x-9 \\ x^2-6x+9=0$

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We'll do this using the fact that:

$9=3^2$ Therefore, we'll represent the rightmost term as a squared term:

$x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-6x+\textcolor{blue}{3}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation we got in the last step:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{3}^2$ indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0$

And indeed it is true that:

$2\cdot x\cdot3=6x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0$

$(x-3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-3=\pm0\\ x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$x^2=6x-9 \\ x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0 \\ \downarrow\\ x-3=0\\ \downarrow\\ \boxed{x=3}$

Therefore the correct answer is answer C.

Answer

$x=3$

Exercise #4

$(x-1)^2=x^2$

Video Solution

Step-by-Step Solution

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$ We'll apply this formula and expand the parentheses in the expressions in the equation:

$(x-1)^2=x^2 \\ x^2-2\cdot x\cdot1+1^2=x^2 \\ x^2-2x+1=x^2 \\$ We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-2x+1=x^2 \\ -2x=-1\hspace{8pt}\text{/}:(-2)\\ \boxed{x=\frac{1}{2}}$ Therefore, the correct answer is answer A.

Answer

$x=\frac{1}{2}$

Exercise #5

$(x-4)^2-x(x+8)=0$

Video Solution

Step-by-Step Solution

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$ We'll apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x-4)^2-x(x+8)=0 \\ x^2-2\cdot x\cdot4+4^2-x^2-8x=0 \\ x^2-8x+16-x^2-8x=0$ In the first stage, we used the distributive property to expand the parentheses,

We'll continue and combine like terms, by moving terms between sides. Then - we can notice that the squared term cancels out and therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-8x+16-x^2-8x=0 \\ -16x=-16\hspace{8pt}\text{/}:(-16)\\ \boxed{x=1}$ Therefore, the correct answer is answer D.