The definition of factoring is
The process of writing a given expression as the product of its factors.
Yes, what factoring means is that we decompose an expression into all the factors that make it up, breaking it down. As shown below:
x^{2} = x \cdot x
Since we have this in mind, we can proceed to mention factoring methods. We will use x and y to refer to the unknowns and a, b and c to refer to the known numbers.
Factorization by grouping of terms
We are going to factor the following expression:
ax^{2} + ay^{2} + bx^{2} + by^{2}
The first thing we will do is group the terms with the unknowns:
\left( ax^{2} + bx^{2}\right) + \left(ay^{2} + by^{2} \right)
Now we will remove the common term from the parentheses:
x^{2} \left( a + b \right) + y^{2} \left( a + b \right)
And we can leave the example here, however, we can factor it a little more. Let’s realize that now the term that we can factor is the (a + b), which our final factorization will be as follows:
(a + b)\left( x^{2} + y^{2} \right)
Factorization by difference of squares
The difference of squares is equal to the product of the sum of two numbers and the subtraction of those numbers, all with exact square roots:
x^{2} - y^{2} = \left( x + y \right) \cdot \left( x - y \right)
Example of factoring by difference of squares
Factor the following expression:
4x^{2} - 81
To perform this exercise we are going to rewrite the equation representing the square root of each term:
4x^{2} - 81 = \left( 2x \right)^{2} - 9^{2}
Now we can write the factorization of the equation, with practice you will skip that step of obtaining the square root of each term:
\left( 2x \right)^{2} - 9^{2} = (2x + 9)(2x - 9)
Factorization of the perfect square trinomial
Many of us already have to know the factorization of this trinomial since we know that it is the result of the perfect square binomial. This trinomial is made up of two terms that are squares and the other term that is the double product of these terms:
x^{2} + 2xy + y^{2} = \left( x + y \right)^{2}
Example of a perfect square trinomial
Factor the following equation:
x^{2} + 10x + 25
To find out if it is a perfect square trinomial we have to make sure that the first and third terms have exact roots, are positive and that the second term is a double product of the square roots.
x^{2} and 25 have exact square roots = x and 5
And now we multiply by two the product of the roots that we just calculated:
2\cdot x \cdot 5 = 10 x
Then the 10x term of our trinomial does correspond to the double product of the square roots.
To write the result, only the square roots are taken and the sum of these roots is squared:
\left( x + 5 \right)^{2}
Example of a perfect square trinomial with negative sign
If it has a negative sign, a perfect square trinomial can still be considered if and only if the negative sign is in the term of the double product of the roots. Let’s see the previous example but now with a negative sign:
x^{2} - 10x + 25
When we have the negative sign, we will have two responses since we, when we square a negative, we will always have a positive. So no matter which of the two terms has the negative in the perfect square binomial, we will always have the same perfect square trinomial, remember the roots of the ends of the trinomial:
x^{2} \rightarrow x \qquad \ 25 \rightarrow 5
Now the double product we add the negative to the x term and then to the 5 term:
2 \cdot (-x) \cdot 5 = - 10 x
2 \cdot x \cdot (-5) = -10 x
So we can have two convenient outcomes if an exercise requires us to apply this negative trinomial perfect square trinomial:
= \left( x - 5\right)^{2} \quad \text{o} \quad \left(5 - x \right)^{2}
Factorization of the trinomial of the form ax^{2} + bx + c
Let’s look at the following example to better understand how this type of factoring works. Factor 2x^{2} + 8x + 6.
Let’s write our trinomial in the following way:
\begin{array}{ c c c c } 2x^{2} & 8x & \ 6 \ & \ \ \\ & & & \\ & & & \\ & & & \end{array}Now, below the extremes of the trinomial (of 2x^{2} and 6), we will write two terms that multiplied give us as a result each of the extremes of our trinomial:
\begin{array}{ c c c c } 2x^{2} & 8x & \ 6 \ & \ \ \\ 2x & & 6 & \\ x & & 1 & \\ & & & \end{array}Okay, the term 2x^{2} can be represented by multiplying 2x\cdot x, and the term 6 can be represented by multiplying 6 \cdot 1. What we will do now is cross multiply each of the terms:
\begin{array}{ c c c c c } 2x^{2} & 8x & \ 6 \ & &\ \ \\ 2x & & 6 & = & 6x \\ x & & 1 & = & 2x \\ & & & & \end{array}As shown in cross multiplication, 2x\cdot 1 = 2x and x \cdot 6 = 6x. We will carry out the sum of the multiplications carried out:
\begin{array}{ c c c c c } 2x^{2} & 8x & \ 6 \ & &\ \ \\ 2x & & 6 & = & 6x \\ x & & 1 & = & 2x \\ & & & = & 8x \end{array}8x = 8x
If the sum is equal to the term in the middle of the perfect square trinomial, that means that we already have our binomials, we will only add the appropriate parentheses with the sign that each term has:
\begin{array}{ c c c c c } 2x^{2} & 8x & \ 6 \ & &\ \ \\ ( 2x & + & 6 )& = & 6x \\ (x & + & 1 )& = & 2x \\ & & & = & 8x \end{array}So the factorization of the trinomial 2x^{2} + 8x + 6 is equal to:
\left( 2x + 6 \right) \left( x + 1 \right)
I am going to tell you that there are trinomials of the form ax^{2} + bx + c that have different factorizations, that is, more answers, look:
\begin{array}{ c c c c c } 2x^{2} & 8x & \ 6 \ & &\ \ \\ 2x & & 2 & = & 2x \\ x & & 3 & = & 6x \\ & & & = & 8x \end{array}We add the appropriate parentheses and signs and observe that another answer of the trinomial 2x2 + 8x + 6 is:
\left( 2x + 2\right) \left( x + 3 \right)
With the same trinomial 2x^{2} + 8x + 6 we will observe that the order in which you arrange the terms may not give you the result:
\begin{array}{ c c c c c } 2x^{2} & 8x & \ 6 \ & &\ \ \\ 2x & & 3 & = & 3x \\ x & & 2 & = & 4x \\ & & & = & 7x \end{array}8x \neq 7x
When that happens to you, simply change the order in which you arrange the multiplications so that when you do the addition, you get the same term in the middle of the trinomial as a result.
Factorization of the sum and difference of perfect cubes
Below we will show you some formulas for the cases of factoring the sum and difference of cubes.
The sum [subtraction] of two perfect cubes is equal to the sum [subtraction] of their cube roots, multiplied by the square of the first cube root, minus [plus] the product of the two cube roots, plus the square of the second Cube root. I leave it more graphic:
x^{3} \pm y^{3} = \left( x \pm y \right) \left(x^{2} \mp xy + y^{2} \right)
I will give you more detailed formulas since the issue of signs is often a bit confused.
Factorization of sum of perfect cubes:
x^{3} + y^{3} = \left( x + y \right) \left(x^{2} - xy + y^{2} \right)
Difference factoring of perfect cubes
x^{3} - y^{3} = \left( x - y \right) \left(x^{2} + xy + y^{2} \right)
Since we have the theory, let’s go with some examples.
Example of the sum of perfect cubes.
Factor the following expression:
8x^{3} + 125
The first thing we have to do is take the cube root of each term:
8x^{3} \rightarrow 2x \qquad \ 125 \rightarrow 5
8x^{3} + 125 = (2x)^{3} + (5)^{3}
Once we have the two terms calculated, let’s take the formula and write the result:
8x^{3} + 125 = \left( 2x + 5 \right) \left( (2x)^{2} – (2x)(5) + (5)^{2} \right)
We carry out the corresponding operations and thus we will obtain our final result:
8x^{3} + 125 = \left( 2x + 5 \right) \left( 4x^{2} – 10x + 25 \right)
Example of the difference of perfect cubes
Factor the following expression:
343y^{3} – 216b^{3}
We take a cube root of each term:
343y^{3} \rightarrow 7y \qquad 216b^{3} \rightarrow 6b
343y^{3} – 216b^{3} = (7y)^{3} – (6b)^{3}
Since we have the two terms calculated, let’s rewrite them in the formula:
343y^{3} – 216b^{3} = \left( 7y – 6b \right) \left( (7y)^{2} + (7y)(6b) + (6b)^{2} \right)
Let’s carry out the corresponding operations to obtain our final result:
343y^{3} – 216b^{3} = \left(7y – 6b \right) \left( 49y^{2} + 42yb + 36b^{2}\right)
Now go out into the world for whatever factoring you have to deal with!
Thank you for being in this moment with us : )
