Short Multiplication Formulas - Examples, Exercises and Solutions

What are Abbreviated Multiplication Formulas?

Abbreviated multiplication formulas, also known as algebraic identities, are shortcuts that simplify the process of expanding and factoring expressions. These formulas save time and reduce the steps required for complex calculations. 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.

Abbreviated Multiplication Formulas for 2nd degree

Here are the basic formulas for abbreviated multiplication:

The square of the sum:
$(X + Y)^2=X^2+ 2XY + Y^2$

The squared difference:

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

The difference of squares:

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

Abbreviated Multiplication Formulas for 3rd degree

Abbreviated multiplication formulas for 3rd-degree expressions, also known as cubic identities, build upon the concepts of the 2nd-degree formulas we’ve already covered. The key difference is the adjustment for working with cubic (3rd-degree) terms instead of quadratic (2nd-degree) terms. These formulas simplify complex cubic expressions, breaking them into manageable parts to make calculations faster and more efficient. They are particularly useful in solving problems involving volumes of cubes and other 3D shapes or in advanced mathematics, such as polynomial factoring and equation solving.

Here are two of the most common abbreviated multiplication formulas for 3rd-degree expressions:

Cube of a Sum

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

Cube of a Difference

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

Abbreviated Multiplication Formulas Verification and Examples

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

Square of a Sum:

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

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

Since: $XY = YX$

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

Example:

$(x+3)^2$

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

Square of a Difference:

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

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

Since: $XY = YX$

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

Example:

$(x−4)^2$

$(x−4)^2=x^2−2(x)(4)+4^2=x^2−8x+16$

Product of a Sum and a Difference:

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

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

Since: $XY = YX$

$- XY + YX = 0$

$X^2 - Y^2$

Example:

$(x+5)(x−5)$

$(x+5)(x−5)=x^2−5^2=x^2−25$

Using Abbreviated Multiplication Formulas to Shift the Expression Both Ways

It’s important to remember that these formulas are not one-sided; you can use them to switch between different forms of an expression as needed. For example, you can use the formula:

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

to expand an expression in parentheses into its expanded form. Conversely, if you encounter an expression like

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

you can factor it back into

$(X + Y)^2$

This flexibility allows you to work with the representation that is most useful for the problem at hand, whether it’s simplifying, solving, or analyzing the expression. Understanding this two-way functionality is essential for mastering algebraic manipulation.

Detailed explanation

Practice Short Multiplication Formulas

Question 1

Solve:

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

Question 2

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

$$ (x+3)^2 $$

Question 3

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

$$ (x-y)^2 $$

Question 4

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

Question 5

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

Examples with solutions for Short Multiplication Formulas

Exercise #1

Solve:

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

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

$4-x^2=0$

We isolate the terms and extract the root:

$4=x^2$

$x=\sqrt{4}$

$x=\pm2$

Answer

±2

Exercise #2

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

$(x+3)^2$

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

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

$x^2+6x+9$

Answer

$x^2+6x+9$

Exercise #3

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 #4

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

Video Solution

Step-by-Step Solution

According to the shortened multiplication formula:

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

$(7+x)^2$

Answer

$(7+x)^2$

Exercise #5

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

Video Solution

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.

Answer

$(2x+5)^2$

Question 1

$$ x^2+144=24x $$

Question 2

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

Question 3

Solve for x:

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

Question 4

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

Question 5

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

Exercise #6

$x^2+144=24x$

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 #7

$x^2=6x-9$

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 #8

Solve for x:

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

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 then apply the formula we mentioned and expand the parentheses in the expression in the equation:

$(x+3)^2=x^2+9 \\ x^2+2\cdot x\cdot3+3^2=x^2+9\\ x^2+6x+9=x^2+9$ We'll continue and combine like terms, by moving terms around. Later - 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:

$x^2+6x+9=x^2+9 \\ 6x=0\hspace{8pt}\text{/}:6\\ \boxed{x=0}$ Therefore, the correct answer is answer A.

Answer

$x=0$

Exercise #9

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

Video Solution

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:

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

$2 x^2-10x+13$

Answer

$2x^2-10x+13$

Exercise #10

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

Video Solution

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:

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

So the correct answer is answer a.

Answer

$y=8$

Question 1

$$ x^2+10x=-25 $$

Question 2

Solve for x:

$$ (x-4)^2=(x+2)(x-1) $$

Question 3

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

Question 4

$$ (x+1)^2=x^2+13 $$

Question 5

$$ (x-1)^2-(x+2)^2=15 $$

Exercise #11

$x^2+10x=-25$

Video Solution

Step-by-Step Solution

Let's solve the given equation:

$x^2+10x=-25$

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

$x^2+10x=-25 \\ x^2+10x+25=0 \\$ Now we notice that the expression on the left side can be factored 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 can do this using the fact that:

$25=5^2$

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

$x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0$

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:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2$ 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 (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 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 as:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it is true that:

$2\cdot x\cdot5=10x$

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

$\textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0$

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

$(x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}$

Let's summarize the solution of the equation:

$x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}$

Therefore the correct answer is answer C.

Answer

$x=-5$

Exercise #12

Solve for x:

$(x-4)^2=(x+2)(x-1)$

Video Solution

Step-by-Step Solution

Let's solve the equation, first we'll simplify the algebraic expressions using the extended distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$ We'll use the shortened multiplication formula for a squared binomial:

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

$(x-4)^2=(x+2)(x-1) \\ x^2-2\cdot x\cdot4+4^2=x^2-x+2x-2 \\ x^2-8x+16=x^2+x-2$ We'll continue and combine like terms, by moving terms between sides - 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:

$x^2-8x+16=x^2+x-2\\ -9x=-18\hspace{8pt}\text{/}:(-9)\\ \boxed{x=2}$ Therefore the correct answer is answer A.

Answer

$x=2$

Exercise #13

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

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 the mentioned formula and expand the parentheses in the expressions in the equation:

$x^2+(x-2)^2=2(x+1)^2 \\ x^2+x^2-2\cdot x\cdot2+2^2=2(x^2+2\cdot x\cdot1+1^2)\\ x^2+x^2-4x+4=2(x^2+2x+1)\\ x^2+x^2-4x+4=2x^2+4x+2\\$ In the final stage, we used the distributive property on the right side of the equation.

We'll continue and combine like terms, by moving terms between sides. Then we can see 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:

$x^2+x^2-4x+4=2x^2+4x+2 \\ -8x=-2\hspace{8pt}\text{/}:(-8)\\ \boxed{x=\frac{2}{8}=\frac{1}{4}}$

In the final stage, we reduced the fraction that was obtained as the solution for $x$ .

Therefore, the correct answer is answer A.

Answer

$x=\frac{1}{4}$

Exercise #14

$(x+1)^2=x^2+13$

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:

$(x+1)^2=x^2+13 \\ x^2+2\cdot x\cdot1+1^2=x^2+13 \\ x^2+2x+1=x^2+13$

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:

$x^2+2x+1=x^2+13 \\ 2x=12\hspace{8pt}\text{/}:2\\ \boxed{x=6}$ Therefore, the correct answer is answer B.

Answer

$x=6$

Exercise #15

$(x-1)^2-(x+2)^2=15$

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:

$(x-1)^2-(x+2)^2=15 \\ x^2-2\cdot x\cdot1+1^2-(x^2+2\cdot x\cdot2+2^2)=15 \\ x^2-2x+1-(x^2+4x+4)=15\\ x^2-2x+1-x^2-4x-4=15$ In the final stage, we used the distributive property to expand the parentheses,

We'll continue and combine like terms, by moving terms between sides, later - 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:

$x^2-2x+1-x^2-4x-4=15 \\ -6x=18\hspace{8pt}\text{/}:(-6)\\ \boxed{x=-3}$ Therefore, the correct answer is answer B.

Answer

$x=-3$

Short Multiplication Formulas - Examples, Exercises and Solutions

What are Abbreviated Multiplication Formulas?

Abbreviated Multiplication Formulas for 2nd degree

Abbreviated Multiplication Formulas for 3rd degree

Abbreviated Multiplication Formulas Verification and Examples

Using Abbreviated Multiplication Formulas to Shift the Expression Both Ways

Practice Short Multiplication Formulas

Examples with solutions for Short Multiplication Formulas

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Video Solution

Step-by-Step Solution

Answer

Exercise #3

Video Solution

Step-by-Step Solution

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Exercise #4

Video Solution

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Exercise #5

Video Solution

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Exercise #6

Video Solution

Step-by-Step Solution

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Exercise #7

Video Solution

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Exercise #8

Video Solution

Step-by-Step Solution

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Exercise #9

Video Solution

Step-by-Step Solution

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Exercise #10

Video Solution

Step-by-Step Solution

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Exercise #11

Video Solution

Step-by-Step Solution

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Exercise #12

Video Solution

Step-by-Step Solution

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Exercise #13

Video Solution

Step-by-Step Solution

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Exercise #14

Video Solution

Step-by-Step Solution

Answer

Exercise #15

Video Solution

Step-by-Step Solution

Answer

More Questions

Square of Difference

Difference of squares

Square of sum