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

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".

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.

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

What is the value of X?

We will look for two numbers whose product is $a\times c$ and whose sum is $b$

We will ask ourselves: which number multiplied by which other will give us $a\times c$ or $c$ (if $a$ equals $1$).

and what plus what would add up to $b$.

In fact, we need to find a pair of numbers that meet these two conditions at the same time.

**We can plot it as follows:**

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$a$ The coefficient of the first term

$b$ The coefficient of the second term

$c$ The constant term

In the first step, we will use only addition to find the first solution, and then, we will use only subtraction to find the second.

Again, the factorization will look as follows:

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

or with subtractions, depending on the solutions.

Test your knowledge

Question 1

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

What is the value of X?

Question 2

\( x^2-12x+36=0 \)

What is the value of X?

Question 3

\( -x^2+13x-14=0 \)

What is the value of X?

$ax^2+bx+c$

The trinomial represents an expression in which $x$ is squared, preceded by a coefficient (which can be positive or negative), but it must not be $0$ (sometimes the coefficient is equal to $1$ and therefore we will not see the $a$), to this term may be added or subtracted some other $bx$ when $b$ represents the coefficient (under the same conditions as $a$) and the independent variable (number $c$) is added or subtracted.

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

We will look for two numbers whose product is $a*c$ and whose sum is $b$

We will ask ourselves: which number multiplied by which other number will give us $a\times c$ or $c$ (if $a$ equals $1$).

and what plus what would add up to $b$.

In fact, we have to find a pair of numbers that meet these two conditions at the same time.

**We can plot it as follows:**

**AC Method:**

We will find all the numbers whose products are $a\times c$ and write them down.

Then, we will see which pair of numbers among those we found will result in $B$.

The two numbers that meet both conditions are the solutions to the trinomial.

**Important**

- If A were different from $1$, it would appear before the parentheses and then there would be a multiplication.
- If any of the solutions or both were negative, we would not add them to the $X$ but subtract them instead.

Do you know what the answer is?

Question 1

\( 2x^2+4x-6=0 \)

What is the value of X?

Question 2

\( x^2-1=0 \)

Question 3

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

$x^2+8x+12$

Let's find all the numbers whose products are $12$ (and remember them in negative as well)

we will obtain:

$12,1$

$2,6$

$3,4$

Now let's see which pair of numbers among those we already found will give us a total of $8$

The pair that meets both conditions is $2,6$.** Let's write the factorization:**

$(x+2)(x+6)$

$x^2+4x+4=$

Let's find our parameters:

$a$ The coefficient of the first term $1$

$b$ The coefficient of the second term $4$

$c$ The constant term $4$

**First, we will place them in the formula with the plus sign and it will give us:**

$\frac{-4+\sqrt{4^2-4\times 1\times 4}}{2\times 1}=$

$\frac{-4+\sqrt{16-16}}{2}=$

$\frac{-4+\sqrt{0}}{2}=$

$\frac{-4}{2}=-2$** We will place them in the formula with the minus sign and we will get:**

$\frac{-4-\sqrt{0}}{2}=$

$-\frac{4}{2}=-2$

We get the same answer.** The factorization is:**

$(x-2)(x-2)$

**If you are interested in this article, you might also be interested in the following articles:**

- Factorization
- The uses of factorization
- Factorization according to short multiplication formulas
- Factorization through the extraction of the common factor outside the parentheses
- Factorization of algebraic fractions
- Addition and subtraction of algebraic fractions
- Simplification of algebraic fractions
- Multiplication and division of algebraic fractions
- Solving equations through factorization

**In the** **Tutorela** **blog, you will find a variety of articles about mathematics.**

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

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-

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

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.__

$x=-3,x=6$

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

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.__

$x=-8,x=-2$

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

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.__

$x=-3,x=6$

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

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.__

$x=2,x=-12$

Check your understanding

Question 1

\( x^2-8x+16=0 \)

Question 2

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

Question 3

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

Related Subjects

- Factorization
- Solving Equations by Factoring
- Extracting the common factor in parentheses
- Factoring using contracted multiplication
- Uses of Factorization
- Factorization: Common factor extraction
- Abbreviated Multiplication Formulas
- The formula for the difference of squares
- The formula for the sum of squares
- Multiplication of the sum of two elements by the difference between them
- Formulas for Cubic Expressions
- The quadratic function
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Finding the Zeros of a Parabola
- Methods for Solving a Quadratic Function
- Squared Trinomial