Factoring Trinomials: Equation factorization

Let's solve the given equation:

$-x^2+13x+14=0$

First, let's arrange the equation, making sure that the coefficient of the quadratic term is positive, we'll do this by multiplying both sides of the equation by $(-1)$:

$-x^2+13x+14=0 \hspace{6pt}\text{/}\cdot(-1) \\ x^2-13x-14=0$

Now we notice that the coefficient of the quadratic term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

We'll 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:

$m\cdot n=-14\\ m+n=-13\\$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 have different signs, according to multiplication rules, and now we'll remember that the possible factors of 14 are 2 and 7 or 14 and 1, fulfilling the second requirement mentioned, together with the fact that the signs of the numbers we're looking for are different from each other leads to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-14\\ n=1 \end{cases}$

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

$x^2-13x-14=0 \\ \downarrow\\ (x-14)(x+1)=0$

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

$x-14=0\\ \boxed{x=14}$

or:

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

Let's summarize the solution of the equation:

$-x^2+13x+14=0 \hspace{6pt}\text{/}\cdot(-1) \\ x^2-13x-14=0\\ \downarrow\\ (x-14)(x+1)=0 \\ \downarrow\\ x-14=0\rightarrow\boxed{x=14}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=14,-1}$

Therefore the correct answer is answer B.

Let's solve the given equation:

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

First, let's simplify the equation, noting that all coefficients and the free term are multiples of the number 2, so we'll divide both sides of the equation by 2:

$2x^2+4x-6=0 \hspace{6pt}\text{/}:2 \\ x^2+2x-3=0\\$Now we notice that 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'll 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:

$m\cdot n=-3\\ m+n=2\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 3 are 3 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other 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:

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

From here we'll remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown on one side:

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

or:

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

Let's summarize then the solution of the equation:

$2x^2+4x-6=0 \\ x^2+2x-3=0 \\ \downarrow\\ (x+3)(x-1)=0 \\ \downarrow\\ x+3=0\rightarrow\boxed{x=-3}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=-3,1}$

Therefore the correct answer is answer B.

Let's try to factor using quick trinomial factoring the given expression:

$x^2+2x+1$

We'll look for a pair of numbers whose product is the free term in the expression, and their sum is the coefficient of the first power term in the expression, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=1\\ m+n=2\\$From the first requirement mentioned, namely - from the multiplication, we should note that the product of the numbers we're looking for needs to yield a positive result, therefore we can conclude that both numbers have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 1 are 1 and -1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

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

and therefore we'll factor the given expression to:

$x^2+2x+1 \\ \downarrow\\ (x+1)(x+1)$

and therefore clearly the factorization suggested in the problem is incorrect.

Therefore - the correct answer is answer B.

Note:

The given question could also be solved by expanding the parentheses in the suggested expression:

$(x+2)(x-1)$ (using the expanded distributive property), and checking if we indeed get the given expression:

$x^2+2x+1$, however it is of course preferable to try to factor the given expression - for practice purposes.

Let's solve the given equation:

$2x^2-16x=-30$

First, let's simplify the equation, noting that all coefficients and the free term are multiples of 2, so we'll divide both sides of the equation by 2, then we'll rearrange it by moving terms:

$2x^2-16x=-30 \hspace{6pt}\text{/}:2 \\ x^2-8x=-15\\ x^2-8x+15=0$

Now we notice that 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'll 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:

$m\cdot n=15\\ m+n=-8\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 15 are 5 and 3 or 15 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have equal signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=-5 \\ n=-3\end{cases}$

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

$x^2-8x+15=0 \\ \downarrow\\ (x-5)(x-3)=0$

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the variable in each:

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

or:

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

Let's summarize then the solution of the equation:

$2x^2-16x=-30\\ x^2-8x+15=0 \\ \downarrow\\ (x-5)(x-3)=0 \\ \downarrow\\ x-5=0\rightarrow\boxed{x=5}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=5,3}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$3x^2-12x=36$

$$First, let's organize the equation by moving terms:

$3x^2-12x=36 \\ 3x^2-12x-36=0$

Note that all coefficients and the free term are multiples of 3, so we'll divide both sides of the equation by 3:

$3x^2-12x-36=0 \hspace{6pt}\text{/}:3 \\ x^2-4x-12=0$

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

Let's 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:

$m\cdot n=-12\\ m+n=-4\\$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 have different signs, according to multiplication rules, and now we'll remember that the possible factors of 12 are 6 and 2, 4 and 3, or 12 and 1. Meeting 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 is:

$\begin{cases}m=-6 \\ n=2\end{cases}$

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

$x^2-4x-12=0 \\ \downarrow\\ (x-6)(x+2)=0$

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the variable in each:

$x-6=0\\ \boxed{x=6}$

or:

$x+2=0\\ \boxed{x=-2}$

Let's summarize the solution of the equation:

$3x^2-12x=36 \\ x^2-4x-12=0 \\ \downarrow\\ (x-6)(x+2)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+2=0\rightarrow\boxed{x=-2}\\ \downarrow\\ \boxed{x=6,-2}$

Therefore the correct answer is answer A.

Let's solve the given equation:

$x^2-10x-4=10-5x$

First, let's organize it by moving terms and combining like terms:

$x^2-10x-4=10-5x \\ x^2-10x-4-10+5x=0 \\ x^2-5x-14=0$

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

Let's 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:

$m\cdot n=-14\\ m+n=-5\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 14 are 7 and 2 or 14 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs leads to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=-7\\ n=2\end{cases}$

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

$x^2-5x-14=0 \\ \downarrow\\ (x-7)(x+2)=0$

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown term:

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

or:

$x+2=0\\ \boxed{x=-2}$

Let's summarize the solution of the equation:

$x^2-10x-4=10-5x \\ x^2-5x-14=0 \\ \downarrow\\ (x-7)(x+2)=0 \\ \downarrow\\ x-7=0\rightarrow\boxed{x=7}\\ x+2=0\rightarrow\boxed{x=-2}\\ \downarrow\\ \boxed{x=7,-2}$

Therefore the correct answer is answer A.

Let's solve the given equation:

$x^2-7x+21=3x$

$$First, let's organize the equation by moving terms and combining like terms:

$x^2-7x+21=3x \\ x^2-7x+21-3x=0\\ x^2-10x+21=0$

Now we notice that 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'll look for a pair of numbers whose product is the constant term and whose sum is the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=21\\ m+n=-10\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 21 are 7 and 3 or 21 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have equal signs, will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases}m=-7 \\ n=-3\end{cases}$

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

$x^2-10x+21=0 \\ \downarrow\\ (x-7)(x-3)=0$

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the variable in each:

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

or:

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

Let's summarize the solution of the equation:

$x^2-7x+21=3x \\ x^2-10x+21=0 \\ \downarrow\\ (x-7)(x-3)=0 \\ \downarrow\\ x-7=0\rightarrow\boxed{x=7}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=7,3}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$2x^2+6x-7=x^2+2x-2$

First, let's organize it by moving terms and combining like terms:

$2x^2+6x-7=x^2+2x-2 \\ 2x^2+6x-7-x^2-2x+2=0 \\ x^2+4x-5=0$

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

Let's 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:

$m\cdot n=-5\\ m+n=4\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 5 are 5 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:

$\begin{cases} m=5\\ n=-1\end{cases}$

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

$x^2+4x-5=0 \\ \downarrow\\ (x+5)(x-1)=0$

From here we'll remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown term:

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

or:

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

Let's summarize then the solution of the equation:

$2x^2+6x-7=x^2+2x-2\\ x^2+4x-5=0 \\ \downarrow\\ (x+5)(x-1)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=-5,1}$

Therefore the correct answer is answer A.