Short Multiplication Formulas - Examples, Exercises and Solutions

We use the abbreviated multiplication formula:

$4-x^2=0$

We isolate the terms and extract the root:

$4=x^2$

$x=\sqrt{4}$

$x=\pm2$

We use the abbreviated multiplication formula:

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

$x^2+6x+9$

We use the abbreviated multiplication formula:

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

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

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

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.

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.

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:

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

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:

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

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.

According to the shortened multiplication formula:

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

$(7+x)^2$

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.

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:

$(x+1)^2=x^2 \\ x^2+2\cdot x\cdot1+1^2=x^2 \\ x^2+2x+1=x^2 \\$We'll continue and combine like terms, by moving terms between sides. 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:

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

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:

$(x+2)^2-12=x^2 \\ x^2+2\cdot x\cdot2+2^2-12=x^2 \\ x^2+4x+4-12=x^2 \\$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:

$x^2+4x+4-12=x^2 \\ 4x=8\hspace{8pt}\text{/}:4\\ \boxed{x=2}$Therefore, the correct answer is answer B.

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.

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.

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.

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.

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

$(x+2)^2=x^2+12 \\ x^2+2\cdot x\cdot2+2^2=x^2+12\\ x^2+4x+4=x^2+12$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:

$x^2+4x+4=x^2+12 \\ 4x=8\hspace{8pt}\text{/}:4\\ \boxed{x=2}$Therefore, the correct answer is answer C.