Factoring Trinomials: Complete the equation

Let's simplify the expression given on the left side:

$(2x+4)(x-5)$ Let's open the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Note that in the formula template for the distribution law mentioned, we take by default that the operation between terms inside the parentheses is addition, so we won't forget of course that the sign preceding a term is an inseparable part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition occurs between all terms (if needed),

Therefore, we'll first present each of the expressions in parentheses in the multiplication on the left side as an expression containing addition:

$(2x+4)(x-5)=2x^2-6x+\textcolor{purple}{\boxed{?}} \\ \downarrow\\ \big(2x+(+4)\big)\big(x+(-5)\big)=2x^2-6x+\textcolor{purple}{\boxed{?}} \\$Now for convenience, let's write again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

And apply it to our problem:

$\big(\textcolor{red}{2x}+\textcolor{blue}{(+4)}\big)\big(x+(-5)\big)=2x^2-6x+\textcolor{purple}{\boxed{?}} \\ \downarrow\\ \textcolor{red}{2x}\cdot x +\textcolor{red}{2x}\cdot (-5)+\textcolor{blue}{(+4)}\cdot x +\textcolor{blue}{(+4)}\cdot (-5)=2x^2-6x+\textcolor{purple}{\boxed{?}} \\$We'll continue and apply the multiplication sign rules, remember that multiplying terms with identical signs yields a positive result, and multiplying terms with different signs yields a negative result, in the next step we'll combine like terms in the expression obtained on the left side:

$\textcolor{red}{2x}\cdot x +\textcolor{red}{2x}\cdot (-5)+\textcolor{blue}{(+4)}\cdot x +\textcolor{blue}{(+4)}\cdot (-5)=2x^2-6x+\textcolor{purple}{\boxed{?}} \\ \downarrow\\ 2x^2-10x+4x-20=2x^2-6x+\textcolor{purple}{\boxed{?}} \\ 2x^2-6x-20=2x^2-6x+\textcolor{purple}{\boxed{?}}$

Therefore, according to the multiplication sign rules, the missing expression is the number $-20$,

In other words - the correct answer is C.

Let's simplify the expression given in the left side:

$(x+3)(x+\textcolor{purple}{\boxed{?}})$ For ease of calculation we will replace the square with the question mark (indicating the missing part that needs to be completed) with the letter $\textcolor{purple}{k}$, meaning we will perform the substitution:

$$$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\$Next, we will expand the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's note that in the formula template for the distribution law mentioned we assume by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the laws of sign multiplication and thus we can represent any expression in parentheses, which we expand using the aforementioned formula, first, as an expression where addition is performed between all terms (if necessary),

Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:

$(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\$Now for convenience, let's write down again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$And we'll apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\$We'll continue and apply the laws of multiplication signs, remembering that multiplying expressions with identical signs will yield a positive result, and multiplying expressions with different signs will yield a negative result:

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\$Now, we want to present the expression on the left side in a form identical to the expression on the right side, that is - as a sum of three terms with different exponents: second power (squared), first power, and zero power (i.e., the free number - not dependent on x). To do this - we will factor out the part of the expression on the left side where the terms are in the first power:

$x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6$

Now in order for equality to hold - we require that the coefficient of the first-power term on both sides of the equation be identical and at the same time - we require that the free term on both sides of the equation be identical as well:

$x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}}$In other words, we require that:

$\begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}$

Let's summarize the solution steps:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\ \downarrow\\ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\ \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases} \\ \textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}}$Therefore, the missing expression is the number $2$meaning - the correct answer is a'.

Let's examine the problem:

$(x-4)(x+\textcolor{purple}{\boxed{?}})=x^2-2x-8$

To complete the missing expression on the left side, we can simply factor into trinomial form (into a product of two binomials) the expression on the right side:

$x^2-2x-8$

Let's proceed with factoring:

Note that in the given expression, the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

We will look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-8\\ m+n=-2\$

From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers must have different signs, according to multiplication rules, and now we'll remember that the possible factors of 8 are 4 and 2 or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-4\\ n=2 \end{cases}$

Therefore we'll factor the expression on the right side to:

$(x-4)(x+\textcolor{purple}{\boxed{?}})=x^2-2x-8 \\ \downarrow\\ (x-4)(x+\textcolor{purple}{\boxed{?}})=(x-4)(x+2)$

Therefore the missing expression is the number 2,

Meaning - the correct answer is answer B.

Let's simplify the expression given on the left side:

$(x+7)(x-2)$ Let's open the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Note that in the formula template for the distribution law mentioned, we take by default that the operation between terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition operation exists between all terms (if needed),

Therefore, we'll first present each of the expressions in parentheses in the multiplication on the left side as an expression with addition operation:

$(x+7)(x-2)=x^2+\textcolor{purple}{\boxed{?}}x-14 \\ \downarrow\\ \big(x+(+7)\big)\big(x+(-2)\big)=x^2+\textcolor{purple}{\boxed{?}}x-14 \\$Now for convenience, let's write again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

And apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+7)}\big)\big(x+(-2)\big)=x^2+\textcolor{purple}{\boxed{?}}x-14 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (-2)+\textcolor{blue}{(+7)}\cdot x +\textcolor{blue}{(+7)}\cdot (-2)=x^2+\textcolor{purple}{\boxed{?}}x-14 \\$We'll continue and apply the multiplication sign rules, remember that multiplying terms with identical signs yields a positive result, and multiplying terms with different signs yields a negative result, in the next step we'll combine like terms in the expression obtained on the left side:

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (-2)+\textcolor{blue}{(+7)}\cdot x +\textcolor{blue}{(+7)}\cdot (-2)=x^2+\textcolor{purple}{\boxed{?}}x-14 \\ \downarrow\\ x^2-2x+7x-14=x^2+\textcolor{purple}{\boxed{?}}x-14 \\ x^2+5x-14=x^2+\textcolor{purple}{\boxed{?}}x-14$

Therefore the missing expression is the number 5,

Meaning - the correct answer is A.

Let's simplify the expression given on the left side:

$(x+7)(x-9)$

Let's open the parentheses using the expanded distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Note that in the formula template for the distribution law mentioned, we take by default that the operation between terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition occurs between all terms (if needed),

Therefore, we'll first present each of the expressions in parentheses in the multiplication on the left side as an expression containing addition:

$(x+7)(x-9)=x^2+\textcolor{purple}{\boxed{?}}x-63 \\ \downarrow\\ \big(x+(+7)\big)\big(x+(-9)\big)=x^2+\textcolor{purple}{\boxed{?}}x-63 \\$Now for convenience, let's write again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

And apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+7)}\big)\big(x+(-9)\big)=x^2+\textcolor{purple}{\boxed{?}}x-63 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (-9)+\textcolor{blue}{(+7)}\cdot x +\textcolor{blue}{(+7)}\cdot (-9)=x^2+\textcolor{purple}{\boxed{?}}x-63 \\$We'll continue and apply the multiplication sign rules, remember that multiplying terms with identical signs yields a positive result, and multiplying terms with different signs yields a negative result, in the next step we'll combine like terms in the expression obtained on the left side:

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (-9)+\textcolor{blue}{(+7)}\cdot x +\textcolor{blue}{(+7)}\cdot (-9)=x^2+\textcolor{purple}{\boxed{?}}x-63 \\ \downarrow\\ x^2-9x+7x-63=x^2+\textcolor{purple}{\boxed{?}}x-63 \\ x^2-2x-63=x^2+\textcolor{purple}{\boxed{?}}x-63$

Therefore the missing expression is the number $-2$,

That is - the correct answer is a'.