Factorization - Examples, Exercises and Solutions

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:

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

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 solve the given equation:

$x^2-12x+36=0$

Note that we can factor the expression on the left side using the perfect square binomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We'll do this using the fact that:

$36=6^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{6}^2=0$

Now let's examine again the perfect square binomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side of the equation that we got in the last step:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{6}^2$indeed match the form of the first and third terms in the perfect square binomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square binomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot6=12x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-6)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-6=\pm0\\ x-6=0\\ \boxed{x=6}$

Let's summarize the solution of the equation:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0 \\ \downarrow\\ x-6=0\\ \downarrow\\ \boxed{x=6}$

Therefore the correct answer is answer A.

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:

$x^2-3x+2=0$

Note 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 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:

$m\cdot n=2\\ m+n=-3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we are looking for must yield a positive result, therefore we can conclude that both numbers have the same sign, according to multiplication rules, and now we'll remember that the possible factors of 2 are 2 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are identical will lead to the conclusion that the only possibility for the two numbers we're looking for is:

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

therefore we will factor the expression on the left side of the equation to:

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

From here we'll remember that the product of 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 on one side:

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

or:

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

Let's summarize then the solution of the equation:

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

Therefore the correct answer is answer B.

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.