Factoring Trinomials

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 solve the given equation:

$x^2-1=0$ We will do this simply by isolating the unknown on one side and taking the square root of both sides:

$x^2-1=0 \\ x^2=1\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=\pm1}$

Therefore, the correct answer is answer A.

Let's observe that the given equation:

$x^2+10x+16=0$is a quadratic equation that can be solved using quick factoring:

$x^2+10x+16=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=16\\ \underline{?}+\underline{?}=10\end{cases}\\ \downarrow\\ (x+2)(x+8)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+2)(x+8)=0 \\ \downarrow\\ x+2=0\rightarrow\boxed{x=-2}\\ x+8=0\rightarrow\boxed{x=-8}\\ \boxed{x=-2,-8}$Therefore, the correct answer is answer B.

Let's observe that the given equation:

$x^2+10x-24=0$is a quadratic equation that can be solved using quick factoring:

$x^2+10x-24=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-24\\ \underline{?}+\underline{?}=10\end{cases}\\ \downarrow\\ (x+12)(x-2)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+12)(x-2)=0 \\ \downarrow\\ x+12=0\rightarrow\boxed{x=-12}\\ x-2=0\rightarrow\boxed{x=2}\\ \boxed{x=-12,2}$Therefore, the correct answer is answer B.