Factoring Trinomials: Solving the problem

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.

Let's observe that the given equation:

$x^2-19x+60=0$is a quadratic equation that can be solved using quick factoring:

$x^2-19x+60=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=60\\ \underline{?}+\underline{?}=-19\end{cases}\\ \downarrow\\ (x-4)(x-15)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-4)(x-15)=0 \\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ x-15=0\rightarrow\boxed{x=15}\\ \boxed{x=4,15}$Therefore, the correct answer is answer A.

Let's observe that the given equation:

$x^2-2x-3=0$is a quadratic equation that can be solved using quick factoring:

$x^2-2x-3=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-3\\ \underline{?}+\underline{?}=-2\end{cases}\\ \downarrow\\ (x-3)(x+1)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-3)(x+1)=0 \\ \downarrow\\ x-3=0\rightarrow\boxed{x=3}\\ x+1=0\rightarrow\boxed{x=-1}\\ \boxed{x=-1,3}$Therefore, the correct answer is answer B.

Let's observe that the given equation:

$x^2-3x-18=0$is a quadratic equation that can be solved using quick factoring:

$x^2-3x-18=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-18\\ \underline{?}+\underline{?}=-3\end{cases}\\ \downarrow\\ (x-6)(x+3)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$Therefore, the correct answer is answer A.

Let's observe that the given equation:

$x^2-3x-18=0$is a quadratic equation that can be solved using quick factoring:

$x^2-3x-18=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-18\\ \underline{?}+\underline{?}=-3\end{cases}\\ \downarrow\\ (x-6)(x+3)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$Therefore, the correct answer is answer A.

Let's observe that the given equation:

$x^2-5x-50=0$is a quadratic equation that can be solved using quick factoring:

$x^2-5x-50=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=-50\\ \underline{?}+\underline{?}=-5\end{cases}\\ \downarrow\\ (x-10)(x+5)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-10)(x+5)=0 \\ \downarrow\\ x-10=0\rightarrow\boxed{x=10}\\ x+5=0\rightarrow\boxed{x=-5}\\ \boxed{x=10,-5}$Therefore, the correct answer is answer C.

Let's observe that the given equation:

$x^2-7x+12=0$is a quadratic equation that can be solved using quick factoring:

$x^2-7x+12=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=12\\ \underline{?}+\underline{?}=-7\end{cases}\\ \downarrow\\ (x-3)(x-4)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x-3)(x-4)=0 \\ \downarrow\\ x-3=0\rightarrow\boxed{x=3}\\ x-4=0\rightarrow\boxed{x=4}\\ \boxed{x=3,4}$Therefore, the correct answer is answer A.

Let's observe that the given equation:

$x^2+9x+20=0$is a quadratic equation that can be solved using quick factoring:

$x^2+9x+20=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=20\\ \underline{?}+\underline{?}=9\end{cases}\\ \downarrow\\ (x+5)(x+4)=0$and therefore we get two simpler equations from which we can extract the solution:

$(x+5)(x+4)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x+4=0\rightarrow\boxed{x=-4}\\ \boxed{x=-4,-5}$Therefore, the correct answer is answer A.

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 solve the given equation:

$x^2+6x+9=0$We can identify that the expression on the left side can be factored using the perfect square trinomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$Let's do this:

$x^2+6x+9=0 \\ x^2\textcolor{blue}{+6x}+3^2=0 \\ x^2\textcolor{blue}{+2\cdot x\cdot3}+3^2=0 \\ \downarrow\\ (x+3)^2=0$We emphasize that factoring using the mentioned formula was possible only because the middle term in the expression (which is in first power in this case and highlighted in blue in the previous calculation) indeed matched the middle term in the perfect square trinomial formula,

We'll continue and solve the resulting equation, which we'll do using square root extraction on both sides:

$(x+3)^2=0 \hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x+3=0\\ \boxed{x=-3}$Therefore, the correct answer is answer B.

Let's solve the given equation:

$x^2-8x+16=0$We identify that we can factor the expression on the left side using the perfect square trinomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$Let's do this:

$x^2-8x+16=0 \\ x^2\textcolor{blue}{-8x}+4^2=0 \\ x^2\textcolor{blue}{-2\cdot x\cdot4}+4^2=0 \\ \downarrow\\ (x-4)^2=0$Note that factoring using this formula was only possible because the middle term in the expression (which is in first power in this case and highlighted in blue in the previous calculation) indeed matched the middle term in the perfect square trinomial formula,

We'll continue and solve the resulting equation by taking the square root of both sides:

$(x-4)^2=0 \hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x-4=0\\ \boxed{x=4}$Therefore, the correct answer is answer C.

Our goal is to factor the expression on the left side of the given equation:

$x^2+x-2=0$

Note that the coefficient of the quadratic term in the expression on the left side is 1, therefore, we can (try to) factor the expression by 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 the given values:

$m\cdot n=-2\\ m+n=1\\$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 the multiplication rules. Note that the possible factors of 2 are 2 and 1, fulfilling the second requirement mentioned. Furthermore the fact that the signs of the numbers are different from each other leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-1\\ n=2 \end{cases}$

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

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

The correct answer is answer A.