Factoring Trinomials: Test if the coefficient is different from 1

Let's solve the given equation:

$3x^2-3x=6$

$$First, let's organize the equation by moving terms and combining like terms:

$3x^2-3x=6 \\ 3x^2-3x-6=0 \\$Note that all coefficients and the free term are multiples of 3, so we'll divide both sides of the equation by 3:

$3x^2-3x-6=0 \hspace{6pt}\text{/}:3 \\ x^2-x-2=0$N

ow 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:

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:

$m\cdot n=-2\\ m+n=-1\\$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 multiplication rules, and now we'll remember that the possible factors of 2 are 2 and 1, satisfying the second requirement mentioned, along with the fact that the numbers we're looking for have different signs 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'll factor the expression on the left side of the equation to:

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

From here we'll remember that the product of expressions equals zero 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 in each:

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

or:

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

Let's summarize the solution of the equation:

$3x^2-3x=6 \\ 3x^2-3x-6=0 \\ x^2-x-2=0 \\ \downarrow\\ (x-2)(x+1)=0 \\ \downarrow\\ x-2=0\rightarrow\boxed{x=2}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=2,-1}$

Therefore the correct answer is answer B.

Let's solve the given equation:

$24x^2-44x=-12$

$$First, let's arrange the equation by moving terms and combining like terms:

$24x^2-44x=-12 \\ 24x^2-44x+12 =0 \\$Now, instead of dividing both sides of the equation by the common factor of all terms in the equation (which is 4), we'll choose to factor it out of the parentheses:

$24x^2-44x+12 =0 \\ 4(6x^2-11x+3)=0$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplying expressions equals zero,

however, the first factor in the expression we got is 4, which is obviously different from zero, therefore:

$6x^2-11x+3=0$

Now we notice that in the resulting equation the coefficient of the quadratic term is not 1, therefore, we'll solve the equation using the quadratic formula (let's recall it):

The rule states that for a quadratic equation in the general form:

$ax^2+bx+c =0$

there are two solutions (or fewer) which we can find using the formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$,

Let's continue and use the quadratic formula, noting that:

$\begin{cases} a=6\\ b=-11\\ c=3 \end{cases}$

therefore the solutions to the quadratic equation are:

$x_{1,2}=\frac{11\pm\sqrt{(-11)^2-4\cdot6\cdot3}}{2\cdot6} \\ x_{1,2}=\frac{11\pm\sqrt{121-72}}{12} \\ x_{1,2}=\frac{11\pm\sqrt{49}}{12}\\ x_{1,2}=\frac{11\pm7}{12}\\ x_1=\frac{11+7}{12}=\frac{18}{12}\\ x_2=\frac{11-7}{12}=\frac{4}{12}\\$Let's continue and reduce the fractions in the solutions we got:

$x_1=\frac{18}{12}=\frac{3\cdot\not{6}}{2\cdot\not{6}}\\ \boxed{x_1=\frac{3}{2}}\\$and

$x_2=\frac{4}{12}=\frac{\not{4}}{3\cdot\not{4}}\\ \boxed{x_2=\frac{1}{3}}\\$

Let's summarize the solution of the equation:

$24x^2-44x=-12 \\ 24x^2-44x+12 =0 \\ \\ \downarrow\\ 4(6x^2-11x+3)=0\\ \downarrow\\ 6x^2-11x+3=0 \\ \downarrow\\ x_{1,2}=\frac{11\pm\sqrt{49}}{12}\\ \downarrow\\ x_1=\frac{18}{12}\rightarrow\boxed{x_1=\frac{3}{2}}\\ x_2=\frac{4}{12}\rightarrow\boxed{x_2=\frac{1}{3}}\\ \downarrow\\ \boxed{x=\frac{3}{2},\frac{1}{3}}$

Therefore, the correct answer is answer D.

Let's solve the given equation:

$5x^2+15x-40=3x+x^2$

$$First, let's organize the equation by moving terms and combining like terms:

$5x^2+15x-40=3x+x^2 \\ 5x^2+15x-40-3x-x^2=0 \\ 4x^2+12x-40=0$

Note that all coefficients and the free term are multiples of 4, so we'll divide both sides of the equation by 4:

$4x^2+12x-40=0\hspace{6pt}\text{/}:4 \\ x^2+3x-10=0$

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:

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:

$m\cdot n=-10\\ m+n=3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must 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 10 are 5 and 2 or 10 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=5 \\ n=-2 \end{cases}$

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

$x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0$

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 variable in each:

$x+5=0\\ \boxed{x=-5}$

or:

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

Let's summarize the solution of the equation:

$5x^2+15x-40=3x+x^2 \\ 4x^2+12x-40=0\\ x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=-5,2}$

Therefore the correct answer is answer A.

Let's solve the given equation:

$4x^2-14x-8=0$

$$Instead of dividing both sides of the equation by the common factor of all terms in the equation (which is the number 2), we will choose to factor it out using parentheses:

$4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0$

Now, remember that multiplying all of the terms in the expression will yield 0 only if at least one of the terms equals zero.

However, the first factor in the expression we got is the number 2, which is obviously not zero, therefore:

$2x^2-7x-4 =0$Now let's note that the coefficient of the quadratic term (squared) is more than 1.

Of course, we can solve the equation using the quadratic formula, but we prefer, for the sake of skill improvement, to continue and factor the expression on the left side.

We will use the grouping method.

Just as in the quick trinomial factoring method (which is actually a special case of the general trinomial factoring method), we will look for a pair of numbers $m,\hspace{2pt}n$whose product give us the product of the coefficient of the quadratic term and the costant term in the general expression:

$ax^2+bx+c$and their sum.

So, we will look for a pair of numbers: $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=a\cdot c\\ m+n=b$ Once we find the pair of numbers that satisfy both conditions mentioned (if indeed such can be found) we will separate the coefficient of the term in the first power accordingly and factor by grouping.

Let's return then to the problem and demonstrate:

In the equation:

$2x^2-7x-4 =0$ We will look for a pair of numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=2\cdot (-4)\\ m+n=-7 \\ \downarrow\\ m\cdot n=-8\\ m+n=-7 \\$We'll continue, just as we do in the quick trinomial factoring method.

From the first requirement mentioned, that is - from the multiplication, let's note that the product of the numbers we're looking for should yield a negative result and therefore we can conclude that the two numbers have different signs, this is according to multiplication laws, and now we'll remember that the possible factors of the number 8 are 2 and 4 or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=1 \end{cases}$From here, unlike the quick trinomial factoring method(where this step is actually skipped and factored directly, but it definitely exists), we will separate the factors of the coefficient according to the pair of numbers $m,\hspace{2pt}n$we found:

$2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{(-8+1)x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\$In the next step we will factor by grouping:

We will refer to two groups of terms, so that in each group there is one term in the first power (the choice of groups doesn't matter - as long as this condition is maintained), in each group - we will factor out a common factor so that inside the parentheses, in both groups, we get the same expression:

$$$\textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x(x-4)}\textcolor{red}{+1(x-4)} =0\\$(In this case, in the second group - which is marked in red, it was not possible to factor further, so we settled for factoring out the number 1 as a common factor for emphasis),

Now, note that the expression in parentheses in both groups is identical and therefore we can refer to it as a common factor for both groups (which is a binomial) and factor it out of the parentheses:

$\textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0$We have thus obtained a factored expression on the left side.

Let's summarize this factoring technique:

$2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\ \downarrow\\ \textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0$This technique is very important - we recommend reviewing it briefly before moving on.

Let's continue solving the equation, we got:

$(x-4)(2x+1)=0$

Here we recall that multiplying the terms in the expression will yield 0 only if at least one of the terms in the expression equals zero.

Therefore we will seperate the two parts of the equation and solve them separately, by isolating the variable:

$x-4=0\\ \boxed{x=4}$Or :

$2x+1=0\\ 2x=-1\text{/}:2\\ \boxed{x=-\frac{1}{2}}$

Let's summarize the solution of the equation:

$4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0 \\ \downarrow\\ 2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\textcolor{blue}{-8x+x}-4 =0 \\ \downarrow\\ 2x\underline{(x-4)}+1\cdot\underline{(x-4)}=0\\ \underline{(x-4)}(2x+1)=0\\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ 2x+1=0\rightarrow\boxed{x=-\frac{1}{2}}\\ \downarrow\\ \boxed{x=4,-\frac{1}{2}}$Therefore the correct answer is answer a.