#### $ax^2+bx+c$

Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".

Question Types:

Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".

$(x+solution \space one)(x+solution\space two)$

or with subtractions, depending on the solutions.

Question 1

\( x^2-1=0 \)

Question 2

\( x^2+10x+16=0 \)

Question 3

\( x^2+10x-24=0 \)

Question 4

\( x^2-19x+60=0 \)

Question 5

\( x^2-2x-3=0 \)

$x^2-1=0$

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.__

$x=\pm1$

$x^2+10x+16=0$

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.__

$x=-8,x=-2$

$x^2+10x-24=0$

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.__

$x=2,x=-12$

$x^2-19x+60=0$

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.__

$x=15,x=4$

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

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.__

$x=3,x=-1$

Question 1

\( x^2-3x-18=0 \)

Question 2

\( x^2-3x-18=0 \)

Question 3

\( x^2-5x-50=0 \)

Question 4

\( x^2+6x+9=0 \)

What is the value of X?

Question 5

\( x^2-7x+12=0 \)

$x^2-3x-18=0$

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.__

$x=-3,x=6$

$x^2-3x-18=0$

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.__

$x=-3,x=6$

$x^2-5x-50=0$

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.__

$x=10,x=-5$

$x^2+6x+9=0$

What is the value of X?

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

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.

3-

$x^2-7x+12=0$

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.__

$x=3,x=4$

Question 1

\( x^2+9x+20=0 \)

Question 2

Complete the equation:

_{\( (x+3)(x+\textcolor{red}{☐})=x^2+5x+6 \)}

Question 3

How many solutions does the equation have?

\( x^2+10x+9=0 \)

Question 4

How many solutions does the equation have?

\( x^4+12x^3+36x^2=0 \)

Question 5

Solve the following equation:

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

$x^2+9x+20=0$

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.__

$x=-4,x=-5$

Complete the equation:

_{$(x+3)(x+\textcolor{red}{☐})=x^2+5x+6$}

**Let's simplify** the expression given in the left side:

$(x+3)(x+\textcolor{purple}{\boxed{?}})$** **__For ease of calculation__ **we will replace** the square with the question mark **(indicating the missing part that needs to be completed)** with the letter $\textcolor{purple}{k}$, meaning we will perform the substitution:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\
\downarrow\\
(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\$Next, **we will expand the parentheses using the expanded distribution law:**

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's note that in the formula template for the distribution law mentioned ** we assume by default that the operation between the terms inside the parentheses is addition**, so we won't forget of course that

Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:

$(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\$Now for convenience, let's write down again the expanded distribution law mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$And we'll apply it to our problem:

$\big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\
\downarrow\\
\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\$We'll continue and apply the ** laws of multiplication signs**, remembering that

$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\
\downarrow\\
x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\$Now, ** we want to present the expression on the left side in a form identical to the expression on the right side**, that is - as a sum of three terms with different exponents: second power (

$x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6$

Now ** in order for equality to hold** - we require that the coefficient of the first-power term

$x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}}$In other words, we require that:

$\begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}$

**Let's summarize** the solution steps:

$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\
\downarrow\\
(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\
\downarrow\\
x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\
\downarrow\\
x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\
\downarrow\\
x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\
\begin{cases}
\textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\
3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2}
\end{cases} \\
\textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}}$__Therefore, the missing expression is__** the number **$2$

2

How many solutions does the equation have?

$x^2+10x+9=0$

Let's observe that the given equation:

$x^2+10x+9=0$is a ** quadratic equation that can be solved using quick factoring**:

$x^2+10x+9=0 \longleftrightarrow\begin{cases}\underline{?}\cdot\underline{?}=9\\
\underline{?}+\underline{?}=10\end{cases}\\
\downarrow\\
(x+1)(x+9)=0$and therefore we get two ** simpler equations** from which we can extract the solution:

$(x+1)(x+9)=0\\
\downarrow\\
x+1=0\rightarrow\boxed{x=-1}\\
x+9=0\rightarrow\boxed{x=-9}\\
\boxed{x=-1,-9}$and therefore the given equation has __two solutions__,

__Thus, the correct answer is answer B.__

Two solutions

How many solutions does the equation have?

$x^4+12x^3+36x^2=0$

Let's solve the given equation:

$x^4+12x^3+36x^2=0$We note that it is possible __to factor the expression__ which is in the left side of the given equation, this is done by __taking out the common factor__ $x^2$ which is the __greatest common factor of the numbers and letters__ in the expression:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0$**We will focus on the** left side of the equation and then on the right side (the number 0).

**Since the only way to get the result 0 from a product is to multiply by 0**, __at least__ one of the expressions in the product on the left side, must be equal to zero,

Meaning:

$x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}$

__Or:__

$x^2+12x+36=0$** I**n order to find the

__Note that the first coefficient is 1__, so we can try to solve it using __the trinomial formula.__

However, we can factor, in this case, also using __the short multiplication formula for a binomial:__

$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$**The reason **for trying factoring in this approach is that we can identify in the left side of the equation we got in the last step, that the two terms which are **in the far sides** (meaning the term in the first position - it is the squared term and the term in the zero position - it is the free number in the expression) can be presented (simply) as a squared term:

$x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0$

Equating the expression on the left side **in the equation:**

$\downarrow\\ x^2+12x+6^2$

To the expression on the right side **in the short formula above:**

$\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

The conclusion from this is that **what remains to check is whether the middle term in the equation** matches the middle term in the short multiplication formula above, meaning - after identifying $a$$b$which are both in the first position in the short multiplication formula above in which $a$and $b$we check if the middle term in the expression in the left side of the equation **can be presented** as $2\cdot a \cdot b$So, we start by presenting the equation of the short formula to the given expression:

$\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2$__And__ ** indeed it holds** that:

$2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x$Meaning **the middle term in the expression in the equation indeed matches the form of the middle term in the short multiplication formula** (highlighted with a line below), mathematically:

$x^2+\underline{12x}+6^2=0 \\
\textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\
\downarrow\\
(\textcolor{red}{x}+\textcolor{blue}{6})^2=0$We can now remember that **a real root can be calculated only for a positive number or for the number zero** (since __it is not possible to get a negative number from squaring a real number itself__), and therefore for an equation there are two real solutions (or one solution) only if:

**Next** we note that if: $(x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\
x+6=0\\
\boxed{x=-6}$then the __only__ solution to the equation is:

$x^4+12x^3+36x^2=0 \\
\downarrow\\
x^2(x^2+12x+36) =0 \\
\downarrow\\
x^2=0\rightarrow\boxed{x=0} \\
x^2+12x+36=0\\
x^2+2\cdot2\cdot6+6^2=0\\
\rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\
\downarrow\\
\boxed{x=0,-6}$Therefore, ** we can summarize what was explained using the following**:

In the quadratic equation:

$ax^2+bx+c =0$in which the coefficients are substituted and the discriminant is calculated:

$a,b$__If it holds:__

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

__There is no (real) solution__ to the equation.

** b.**$\Delta$:

__There exists a single (real) solution__ to the equation.

** c.**$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$:

__There exist two (real) solutions__ to the equation.

** Now let's return to the given equation** and extract from it the

$\Delta\geq0$We continue and __calculate $\Delta=0$:__

$x=-\frac{b}{2a}$Therefore for the quadratic equation that we solved, ** one (real) solution**,

and in combination with the solution $ax^2+bx+c =0$(the additional solution we found for the given equation which is indicated in the first step after factoring using the common factor),

**Therefore we get that for the given equation:**

$\Delta=b^2-4ac$

__two real solutions.__

Two solutions

Solve the following equation:

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

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

$\begin{cases}
m=-8\\
n=1
\end{cases}$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.__

$4,-\frac{1}{2}$