Factoring Trinomials - Examples, Exercises and Solutions

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.

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 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 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 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 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.

$$The equation in the problem is:

$x^2+6x+9=0$We want to solve this equation using factoring,

First, we'll check if we can factor out a common factor, but this is not possible, since there is no common factor for all three terms on the left side of the equation, we can identify that we can factor the expression on the left side using the quadratic formula for a trinomial squared, however, we prefer to factor it using the factoring method according to trinomials, let's refer to the search for Factoring by trinomials:

Let's note that the coefficient of the squared term (the term with the second power) is 1, so we can try to perform factoring according to the quick trinomial method: (This factoring is also called "automatic trinomial"),

But before we do this in the problem - let's recall the general rule for factoring by quick trinomial method:

The rule states that for the algebraic quadratic expression of the general form:

$x^2+bx+c$We can find a factorization in the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the following conditions are met (conditions of the quick trinomial method):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$which is its factored form (product factors) of the expression,

Let's return now to the equation in the problem that we received in the last stage after arranging it:

$x^2+6x+9=0$Let's note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=9 \\ b=6 \end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:

$\begin{cases} m\cdot n=9\\ m+n=6 \end{cases}$We'll try to identify this pair of numbers through logical thinking and using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:

$m\cdot n=9$We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical,

Next, we'll consider the factors (integers) of the number 9, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 3 and 3, or 9 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities regarding the fulfillment of the second condition:

$m+n=6$ will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:

$3,\hspace{4pt}3$That is - for:

$m=3,\hspace{4pt}n=3$(It doesn't matter which one we call m and which one we call n)

It is satisfied that:

$\begin{cases} \underline{3}\cdot \underline{3}=9\\ \underline{3}+\underline{3}=6 \end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+6x+9 \\ \downarrow\\ (x+3)(x+3)$

In other words, we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

If so, we have factored the quadratic expression on the left side of the equation into factors using the quick trinomial method, and the equation is:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\$$$$$Where in the last stage we noticed that in the expression on the left side the term:

$(x+3)$

multiplies itself and therefore the expression can be written as a squared term:

$(x+3)^2$

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to quickly solve the equation we received:

$(x+3)^2=0$

Let's pay attention to a simple fact, on the left side there is a term raised to the second power, and on the right side the number 0,

and only 0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

$x+3=0$(In the same way we could have operated algebraically in a pure form and taken the square root of both sides of the equation, we'll discuss this in a note at the end)

We'll solve this equation by moving the free number to the other side and we'll get that the only solution is:

$x=-3$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method, we got that:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ \downarrow\\ x+3=0\\ x=-3$Therefore, the correct answer is answer C.

Note:

We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

$(x+3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+3)^2}=\pm\sqrt{0} \\ x+3=\pm0\\ x+3=0$Where on the left side the root (which is a half power) and the second power canceled each other out (this follows from the law of powers for power over power), and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached with logical and unambiguous thinking earlier - in the solution above,

In any other case where on the right side was a number different from 0, we could have solved only by taking the root etc. and considering the two positive and negative possibilities which would then give two different possibilities for the solution.

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.