Factoring Trinomials

I present to you the following trinomial

$ax^2+bx+c$

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".

The factorization will look like this:

$(x+solution \space one)(x+solution\space two)$
or with subtractions, depending on the solutions.

Detailed explanation

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Test yourself on factoring trinomials!

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

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The first way to factor a trinomial

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:

The second way to factor a trinomial - quadratic formula

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

$a$ 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.

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Test your knowledge

Question 1

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

Question 2

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

Question 3

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

What is a trinomial?

$ax^2+bx+c$

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".

The first way to factor a 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

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

Question 2

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

Question 3

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

Let's look at an example of the use of factoring trinomials in the first way.

$x^2+8x+12$
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:
$(x+2)(x+6)$

The second way to factor a trinomial

Let's look at an example of the use of factoring trinomials in the second way:

$x^2+4x+4=$

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

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Examples and exercises with solutions for factoring trinomials

Exercise #1

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

Video Solution

Step-by-Step Solution

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.

Answer

$x=-3,x=6$

Exercise #2

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

Video Solution

Step-by-Step Solution

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.

Answer

$x=-8,x=-2$

Exercise #3

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

Video Solution

Step-by-Step Solution

Let's observe that the given equation:

$x^2-3x-18=0$ is a quadratic equation that can be solved using quick factoring:

$(x-6)(x+3)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+3=0\rightarrow\boxed{x=-3}\\ \boxed{x=6,-3}$ Therefore, the correct answer is answer A.

Answer

$x=-3,x=6$

Exercise #4

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

Video Solution

Step-by-Step Solution

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.

Answer

$x=2,x=-12$

Exercise #5

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

Video Solution

Step-by-Step Solution

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.

Answer

$x=15,x=4$

Check your understanding

Question 1

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

Question 2

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

Question 3

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

Start practice

Factoring Trinomials

I present to you the following trinomial

ax2+bx+cax^2+bx+cax2+bx+c

The factorization will look like this:

Test yourself on factoring trinomials!

The first way to factor a trinomial

The second way to factor a trinomial - quadratic formula

What is a trinomial?

The first way to factor a trinomial

Let's look at an example of the use of factoring trinomials in the first way.

The second way to factor a trinomial

Let's look at an example of the use of factoring trinomials in the second way:

Examples and exercises with solutions for factoring trinomials

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Video Solution

Step-by-Step Solution

Answer

Exercise #3

Video Solution

Step-by-Step Solution

Answer

Exercise #4

Video Solution

Step-by-Step Solution

Answer

Exercise #5

Video Solution

Step-by-Step Solution

Answer

More Questions

Factoring Trinomials

$ax^2+bx+c$