Factoring Trinomials: Using short multiplication formulas

First, let's recall the formula for calculating the area of a square with side length y (length units):

$S_{\boxed{}}=y^2$Therefore, for a square with side length:

$x+1$ (length units), the expression for the area is:

$$$S_{\boxed{}}=(x+1)^2$Now, in order to simplify the expression, let's recall the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$Let's continue and apply this formula to the area expression we got:

$S_{\boxed{}}=(x+1)^2 \\ \downarrow\\ S_{\boxed{}}=x^2+2\cdot x\cdot 1+1\\ \boxed{ S_{\boxed{}}=x^2+2x+1}\\$This is the most simplified expression for the given square's area,

therefore the correct answer is answer D.

First, recall the formula for calculating the area of a rectangle with sides of length a,b (length units):

$S_{\boxed{\hspace{8pt}}}=a\cdot b$

Therefore, by direct calculation, for the rectangle shown in the drawing with side lengths:

$x+1,\hspace{6pt}x-4$ (length units),

The expression for the area is:

$$$S_{\boxed{\hspace{8pt}}}=(x+1)(x-4)$

However, from the given information, we know that the expression for the area of the rectangle in the drawing is:

$S_{\boxed{\hspace{8pt}}}=x^2-13$

Therefore, we can conclude the existence of the equation:

$(x+1)(x-4)=x^2-13$

Now, in order to simplify the equation, recall the expanded distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$

Proceed to solve the equation that we obtained. First, we'll open the parentheses on the left side, then we'll move and combine like terms, and solve the resulting simple equation:

$(x+1)(x-4)=x^2-13 \\ x^2-4x+x-4=x^2-13\\ -3x=-10\hspace{9pt}\text{/:(-3)}\\ \boxed{x=3}$(length units),

Note- this solution for the unknown does not contradict the domain of definition (where the side lengths must be positive, as required) and the area obtained by substituting it into the given expression for the area in the problem:

$S_{\boxed{\hspace{8pt}}}=x^2-13 \\ \downarrow\\ S_{\boxed{\hspace{8pt}}}=5^2-13 =25-13=\boxed{12}$(area units)

Indeed positive, as expected.

Therefore, the correct answer is answer C.

First, let's recall the formula for calculating the area of a square with side length y (length units):

$A_{\boxed{}}=y^2$

Therefore, for a square with side length:

$x+1$

(length units), the expression for area is:

$$$A_{\boxed{}}=(x+1)^2$ (sq. units)

However the given data states that the square's area is 36 sq. units, meaning, in mathematical notation:

$A_{\boxed{}}=36$ (sq. units)

Therefore we can deduce the equation for the unknown x:

$(x+1)^2=36$

Let's continue and solve the resulting equation, starting by simplifying the expression on the left side,

To simplify the expression let's recall the formula for the square of a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

Let's continue and apply this formula to our equation, then combine like terms:

$(x+1)^2=36 \\ \downarrow\\ x^2+2\cdot x\cdot 1+1=36\\ x^2+2x-35=0$

We have obtained a quadratic equation, we can identify that the coefficient of the squared term is 1, therefore we can (try to) solve it using the quick factoring method,

We'll look for a pair of numbers whose product equals the constant term on the left side, and whose sum equals the coefficient of the first-degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-35\\ m+n=2$

From the first requirement above, meaning the multiplication, we can conclude according to the rules of sign multiplication that the two numbers have opposite signs, and now we'll remember that the possible factor pairs of 35 are the numbers 7 and 5 or 35 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have opposite signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=7\\ n=-5 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+2x-35=0 \\ \downarrow\\ (x+7)(x-5)=0$

Where we used the pair of numbers we found earlier in this factoring,

We'll continue and consider the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x+7=0\\ \boxed{x=-7}$

Or:

$x-5=0\\ \boxed{x=5}$

However, from the domain of definition for x specified in the problem (which comes from the fact that the length of a side is positive):

x>-1 We can eliminate the solution: $x=-7$Therefore the only solution to the unknown in the problem that satisfies the given data is:

$\boxed{x=5}$

Therefore the correct answer is answer A.

In order to find the unknown in the problem, let's first recall the Pythagorean theorem which states that the sum of squares of the legs in a right triangle (the sides containing the right angle) equals the square of the hypotenuse (the side opposite to the right angle),

In other words, mathematically,

in a right triangle with legs of length: $a,b$ and hypotenuse of length:

$c$it is always true that:

$a^2+b^2=c^2$

Let's return then to the triangle given in the problem, from the triangle drawing we notice that the legs' lengths are:

$x,\hspace{2pt}x+1$

and the hypotenuse length is:

$x+2$

Therefore, according to the Pythagorean theorem we have:

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

Let's continue and solve the resulting equation, we'll start by simplifying the expressions on both sides,

For this we'll recall the square of binomial formula:

$(z\pm w)^2=z^2\pm2zw+w^2$

Let's apply this formula to the equation we got, first let's expand the parentheses, then combine like terms:

$x^2+\textcolor{red}{(x+1)^2}=\textcolor{blue}{(x+2)^2}\\ \downarrow\\ x^2+\textcolor{red}{x^2+2\cdot x\cdot1+1^2}=\textcolor{blue}{x^2+2\cdot x\cdot2+2^2}\\ x^2+x^2+2x+1=x^2+4x+4\\ x^2-2x-3=0$

We've therefore obtained a quadratic equation, we identify that the coefficient of the quadratic term is 1, so we can (try to) solve it using the quick factoring method,

Let's look for a pair of numbers whose product equals the constant term on the left side, and whose sum equals the coefficient of the first degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-3\\ m+n=-2$

From the first requirement above, namely the product, we can deduce according to the rules of sign multiplication that the two numbers have opposite signs, and now we'll remember that the only possible pair of factors of the (prime) number 3 are 3 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have opposite signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-3\\ n=1 \end{cases}$

Therefore we can factor the expression on the left side of the equation to:

$x^2-2x-3=0 \\ \downarrow\\ (x-3)(x+1)=0$

Where we used the pair of numbers we found earlier in this factorization,

Let's continue and consider the fact that on the left side of the last equation we have a product of algebraic expressions and on the right side we have 0, therefore, since the only way to get a product of 0 is to multiply by 0, at least one of the expressions in the product on the left side must equal zero,

Meaning:

$x-3=0\\ \boxed{x=3}$

Or:

$x+1=0\\ \boxed{x=-1}$

However, from the domain of definition for x specified in the problem:

x>1

We can eliminate the solution: $x=-1$Therefore the only solution to the unknown in the problem that satisfies the given condition is:

$\boxed{x=3}$

Therefore the correct answer is answer A.

In order to find the unknown in the problem, let's first recall the Pythagorean theorem which states that the sum of squares of the legs in a right triangle (the sides containing the right angle) equals the square of the hypotenuse (the side opposite to the right angle),

In other words, mathematically,

in a right triangle with legs of length: $a,b$ and hypotenuse of length:

$c$it is always true that:

$a^2+b^2=c^2$

Let's return then to the triangle given in the problem, from the triangle's drawing we notice that the lengths of its legs are:

$x,\hspace{2pt}x-1$

and the length of the hypotenuse is:

$x+2$

Therefore, according to the Pythagorean theorem we have:

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

Let's continue and solve the resulting equation, we'll start by simplifying the expressions on both sides,

For this we'll recall the shortened multiplication formula for squaring a binomial:

$(z\pm w)^2=z^2\pm2zw+w^2$

Let's apply this formula to the equation we got, first let's expand the parentheses, then combine like terms:

$x^2+\textcolor{red}{(x-1)^2}=\textcolor{blue}{(x+2)^2}\\ \downarrow\\ x^2+\textcolor{red}{x^2-2\cdot x\cdot1+1^2}=\textcolor{blue}{x^2+2\cdot x\cdot2+2^2}\\ x^2+x^2-2x+1=x^2+4x+4\\ x^2-6x-3=0$

(We'll stop here for now, continuation after data modification and correction)

We have therefore obtained a quadratic equation, we identify that the coefficient of the squared term is 1, so we can (try to) solve it using the quick trinomial method,

We'll look for a pair of numbers whose product is the free term on the left side, and whose sum is the coefficient of the first-degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-3\\ m+n=-2$

From the first requirement above, namely the multiplication, we can deduce according to the rules of sign multiplication that the two numbers have different signs, and now we'll remember that the only possible pair of factors of the (prime) number 3 are 3 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-3\\ n=1 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2-2x-3=0 \\ \downarrow\\ (x-3)(x+1)=0$

where we used the pair of numbers we found earlier in this factorization,

We'll continue and consider the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x-3=0\\ \boxed{x=3}$

Or:

$x+1=0\\ \boxed{x=-1}$

However, from the domain of definition for x specified in the problem:

x>1 We can eliminate the solution: $x=-1$

therefore the only solution to the unknown in the problem that satisfies the given is:

$\boxed{x=3}$

Therefore the correct answer is answer A.

In order to find the unknown in the problem, let's first recall the Pythagorean theorem which states that the sum of squares of the legs in a right triangle (the sides containing the right angle) equals the square of the hypotenuse (the side opposite to the right angle),

In other words, mathematically,

in a right triangle with legs of length: $a,b$ and hypotenuse of length:

$c$it is always true that:

$a^2+b^2=c^2$

Let's return then to the triangle given in the problem, from the triangle's drawing we notice that the lengths of its legs are:

$x,\hspace{2pt}x+7$

and the length of the hypotenuse is:

$13$

Therefore, according to the Pythagorean theorem we have:

$x^2+(x+7)^2=13^2$

Let's continue and solve the resulting equation, we'll start by simplifying the expressions on both sides,

For this, let's recall the perfect square binomial formula:

$(z\pm w)^2=z^2\pm2zw+w^2$

Let's apply this formula to the equation we got, first let's expand the parentheses, then combine like terms:

$x^2+\textcolor{red}{(x+7)^2}=13^2\\ \downarrow\\ x^2+\textcolor{red}{x^2+2\cdot x\cdot7+7^2}=13^2\\ x^2+x^2+14x+49=169\\ 2x^2+14x-120=0 \hspace{6pt}\text{/:2}\\ x^2+7x-60=0$

In the final stage we identified that we can further simplify the equation by dividing both sides by 2, since all coefficients in the equation are divisible by 2 with no remainder,

We therefore got a quadratic equation, we identify that the coefficient of the quadratic term is 1, so we can (try to) solve it using the quick factoring method,

Let's look for a pair of numbers whose product is the free term on the left side of the equation, and whose sum is the coefficient of the first-degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-60\\ m+n=7$

From the first requirement above, namely the multiplication, we can deduce according to the rules of sign multiplication that the two numbers have different signs, and now noting that 60 has several pairs of integer factors, including the pair 12 and 5 (we won't list them all here), satisfying the second requirement mentioned, together with the fact that the numbers we're looking for have opposite signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=12\\ n=-5 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+7x-60=0 \\ \downarrow\\ (x+12)(x-5)=0$

where we used the pair of numbers we found earlier in this factorization,

Let's continue and consider the fact that on the left side of the resulting equation we have a product of algebraic expressions and on the right side we have 0, therefore, since the only way to get a product of 0 is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x+12=0\\ \boxed{x=-12}$

Or:

$x-5=0\\ \boxed{x=5}$

However, from the domain of definition for x specified in the problem:

x>3

We can eliminate the solution: $x=-12$

Therefore the only solution to the unknown in the problem that satisfies the given is:

$\boxed{x=5}$

Therefore the correct answer is answer A.