Abbreviated multiplication formulas will be used throughout our math studies, from elementary school to high school. In many cases, we will need to know how to expand or add these equations to arrive at the solution to various math exercises.

Just like other math topics, even in the case of abbreviated multiplication formulas, there is nothing to fear. Understanding the formulas and lots of practice on the topic will give you complete control. So let's get started :)

## Abbreviated Multiplication Formulas for 2nd Grade

Here are the basic formulas for abbreviated multiplication:

$(X + Y)^2=X^2+ 2XY + Y^2$

$(X - Y)^2=X^2 - 2XY + Y^2$

$(X + Y)\times (X - Y) = X^2 - Y^2$

## Abbreviated Multiplication Formulas for 3rd Grade

$(a+b)^3=a^3+3a^2 b+3ab^2+b^3$

$​​​​​​​(a-b)^3=a^3-3a^2 b+3ab^2-b^3$

## Abbreviated Multiplication Formulas Verification

We will test the shortcut multiplication formulas by expanding the parentheses.

$(X + Y)^2 = (X + Y)\times (X+Y) =$

$X^2 + XY + YX + Y^2=$

Since: $XY = YX$

$X^2 + 2XY + Y^2$

$(X - Y)^2 = (X - Y)\times (X-Y) =$

$X^2 - XY - YX + Y^2=$

Since:$XY = YX$

$X^2 - 2XY + Y^2$

$(X + Y)\times (X-Y) =$

$X^2 - XY + YX - Y^2=$

Since: $XY = YX$

$- XY + YX = 0$

$X^2 - Y^2$

## Abbreviated Multiplication Practice

$(X + 2)^2=X^2 - 8$

$-(X + 2)^2=-X^2 - 8$

$(X + 3)^2=(X-4)\times (X+4)$

`

## Abbreviated Multiplication Practice Solutions

$(X + 2)^2=X^2 - 8$

$-(X + 2)^2=-X^2 - 8$

$(X + 3)^2=(X-4)\times (X+4)$

## Examples with solutions for Short Multiplication Formulas

### Exercise #1

Solve:

$(2+x)(2-x)=0$

### Step-by-Step Solution

We use the abbreviated multiplication formula:

$4-x^2=0$

We isolate the terms and extract the root:

$4=x^2$

$x=\sqrt{4}$

$x=\pm2$

±2

### Exercise #2

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

$(x+3)^2$

### Step-by-Step Solution

We use the abbreviated multiplication formula:

$x^2+2\times x\times3+3^2=$

$x^2+6x+9$

$x^2+6x+9$

### Exercise #3

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

$(x-y)^2$

### Step-by-Step Solution

We use the abbreviated multiplication formula:

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

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

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

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

### Exercise #4

$4x^2+20x+25=$

### Step-by-Step Solution

In this task, we are asked to simplify the formula using the abbreviated multiplication formulas.

Let's take a look at the formulas:

$(x-y)^2=x^2-2xy+y^2$

$(x+y)^2=x^2+2xy+y^2$

$(x+y)\times(x-y)=x^2-y^2$

Taking into account that in the given exercise there is only addition operation, the appropriate formula is the second one:

Now let us consider, what number when multiplied by itself will equal 4 and what number when multiplied by itself will equal 25?

The answers are respectively 2 and 5:

We insert these into the formula:

$(2x+5)^2=$

$(2x+5)(2x+5)=$

$2x\times2x+2x\times5+2x\times5+5\times5=$

$4x^2+20x+25$

That means our solution is correct.

$(2x+5)^2$

### Exercise #5

$(7+x)(7+x)=\text{?}$

### Step-by-Step Solution

According to the shortened multiplication formula:

Since 7 and X appear twice, we raise both terms to the power:

$(7+x)^2$

$(7+x)^2$

### Exercise #6

$(x-2)^2+(x-3)^2=$

### Step-by-Step Solution

In order to solve the question, we need to know one of the shortcut multiplication formulas:

$(x−y)^2=x^2−2xy+y^2$

We apply the formula twice:

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

$(x-3)^2=x^2-6x+9$

Now we add the two together:

$x^2-4x+4+x^2-6x+9=$

$2 x^2-10x+13$

$2x^2-10x+13$

### Exercise #7

$(2\lbrack x+3\rbrack)^2=$

### Step-by-Step Solution

We will first solve the exercise by opening the inner brackets:

(2[x+3])²

(2x+6)²

We will then use the shortcut multiplication formula:

(X+Y)²=+2XY+

(2x+6)² = 2x² + 2x*6*2 + 6² = 2x+24x+36

$4x^2+24x+36$

### Exercise #8

$2(x+3)^2+3(x+2)^2=$

### Step-by-Step Solution

In order to solve the exercise, remember the abbreviated multiplication formulas:

$(x+y)^2=x^2+2xy+y^2$

Let's start by using the property in both cases:

$(x+3)^2=x^2+6x+9$

$(x+2)^2=x^2+4x+4$

We then reinsert them back into the formula as follows:

$2(x^2+6x+9)+3(x^2+4x+4)=$

$2x^2+12x+18+3x^2+12x+12=$

$5x^2+24x+30$

$5x^2+24x+30$

### Exercise #9

$(2x)^2-3=6$

### Step-by-Step Solution

First we rearrange the equation and set it to 0

$4x^2-3-6=0$

$4x^2-9=0$

We then apply the shortcut multiplication formula:

$4(x^2-\frac{9}{4})=0$

$x^2-(\frac{3}{2})^2=0$

$(x-\frac{3}{2})(x+\frac{3}{2})=0$

$x=\pm\frac{3}{2}$

$±\frac{3}{2}$

### Exercise #10

Solve the following equation:

$x^2+10x+50=-4x+1$

### Step-by-Step Solution

The equation in the problem is:

$x^2+10x+50=-4x+1$First, we identify that the equation is quadratic (and this is because the quadratic term in it does not cancel out), therefore, we will simplify the equation by moving all terms to one side and combine thelike terms:

$x^2+10x+50=-4x+1 \\ x^2+10x+4x+50-1 =0 \\ x^2+14x+49 =0 \\$

We want to solve this equation using factorization.

First, we'll check if we can find a common factor, but this is not possible, since there is no multiplicative factor common to all three terms on the left side of the equation.

We can factor the expression on the left side using the quadratic factoring formula for a trinomial, however, we prefer to factor it using the trinomial factoring methodl:

Note that the coefficient of the quadratic term (the term with the second power) is 1, and therefore we can try to perform factoring according to the quick trinomial method:

But before we do this in the problem - let's remember the general rule for factoring with the quick trinomial method:

The rule states that for the algebraic quadratic expression:

$x^2+bx+c$We can find a factorization to the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the conditions (conditions of the quick trinomial method) are met:

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$which is its factored form (product factors) of the expression,

Let's return now to the equation in the problem that we received in the last stage after arranging it:

$x^2+14x+49 =0$Note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=49 \\ b=14 \end{cases}$Don't forget to consider the coefficient together with its sign.

Let's continue - we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:

$\begin{cases} m\cdot n=49\\ m+n=14 \end{cases}$We'll try to identify this pair of numbers using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:

$m\cdot n=49$We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical.

Next, we'll refer to the factors (integers) of the number 49, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 7 and 7, or 49 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities for the second condition:

$m+n=14$ will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:

$7,\hspace{4pt}7$That is:

$m=7,\hspace{4pt}n=7$(It doesn't matter which one we call m and which one we call n)

It is satisfied that:

$\begin{cases} \underline{7}\cdot \underline{7}=49\\ \underline{7}+\underline{7}=14 \end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+14x+49 \\ \downarrow\\ (x+7)(x+7)$

In other words, we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

If so we factored the quadratic expression on the left side of the equation into factors using factoring according to the quick trinomial method, and the equation is:

$x^2+14x+49=0 \\ \downarrow\\ (x+7)(x+7)=0\\ (x+7)^2=0\\$In the last stage we notice that the expression on the left side the term:

$(x+7)$

is multiplied by itself and therefore the expression can be written as a squared term:

$(x+7)^2$

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to the quick solution of the equation we received:

$(x+7)^2=0$

Let's pay attention to a simple fact, on the left side there is a term that is raised to the second power, and on the right side the number 0.

0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

$x+7=0$(We could have solved algebraically and taken the square root of both sides of the equation, we'll discuss this in a note at the end)

We'll solve this equation by transferring the constant number to the other side and we'll get that the only solution is:

$x=-7$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method:

$x^2+14x+49=0 \\ \downarrow\\ (x+7)(x+7)=0\\ (x+7)^2=0\\ \downarrow\\ x+7=0\\ x=-7$Therefore, the correct answer is answer B.

Note:

We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

$(x+7)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+7)^2}=\pm\sqrt{0} \\ x+7=\pm0\\ x+7=0$

On the left side, the root (which is a half power) and the second power canceled each other out, and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached by logic - in the solution above.

In a case where on the right side there's a number other than 0, we could solve only by taking the root and considering the two positive and negative possibilities which would then give two different possibilities for the solution.

$x=-7$

### Exercise #11

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

### Step-by-Step Solution

In order to solve the exercise, we will need to know the abbreviated multiplication formula:

In this exercise, we will use the formula twice:

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

$(x+2)^2=x^2+4x+4$

$x^2+2x+1+x^2+4x+4=2x^2+6x+5$

x²+2x+1+x²+4x+4=
2x²+6x+5

Note that a common factor can be extracted from part of the digits: $2(x^2+3x)+5$

$2(x^2+3x)+5$

### Exercise #12

$60-16y+y^2=-4$

### Step-by-Step Solution

Let's solve the given equation:

$60-16y+y^2=-4$First, let's arrange the equation by moving terms:

$60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$Now let's examine again the short factoring formula we 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 of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, 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 it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$And indeed it holds that:

$2\cdot y\cdot8=16y$So we can present the expression on the left side of the given equation as a difference of two squares:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

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

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

$y=8$

### Exercise #13

Look at the following square:

Express the area of the square.

### Step-by-Step Solution

Remember that the area of a square is equal to the side of the square squared

The formula for the area of a square is:

$S=a^2$

Now let's substitute the data into the formula:

$S=(5+2x)^2$

$(5+2x)^2$

### Exercise #14

Look at the square below:

Express its area in terms of $x$.

### Step-by-Step Solution

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

The formula for the area of the square is

$A=L^2$

We place the data in the formula:

$A=(x-7)^2$

$(x-7)^2$

### Exercise #15

Given the rectangle ABCD

AB=X

The ratio between AB and BC is $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

### Step-by-Step Solution

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$

$m^2+1=(x+1)^2$