Factorization Practice Problems - Master Algebraic Techniques

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.