In mathematics, for factorization we intend to find the numbers or expressions that by multiplying each other give a certain number or equation. Factoring is a useful skill to learn in solving algebraic problems; when dealing with second degree equations or other types of polynomials, the ability to factorize becomes almost essential. Factorization can be used to simplify algebraic expressions and facilitate calculations. It also allows you to eliminate some results faster than the classic resolution.
Steps
Method 1 of 3: Factoring Simple Numbers and Algebraic Expressions
Step 1. Understand the definition of factoring applied to single numbers
Factorization is theoretically simple, but in practice it can be challenging when applied to complex equations. This is why it is easier to approach factorization starting with simple numbers and then moving on to simple equations and then to more complex applications. The factors of a certain number are the numbers that multiplied together produce that number. For example, the factors of 12 are 1, 12, 2, 6, 3, and 4, because 1 × 12, 2 × 6, and 3 × 4 all make 12.
- Another way to think about it is that the factors of a given number are the numbers that exactly divide that number.
-
Can you spot all the factors of the number 60? The number 60 is used for many purposes (minutes in an hour, seconds in a minute, etc.) because it is exactly divisible by many numbers.
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60
Step 2. Note that expressions that contain unknowns can also be divided into factors
Just like single numbers, unknowns with numerical coefficients (monomials) can also be factored. To do this, just find the factors of the coefficient. Knowing how to factor monomials is useful for simplifying the algebraic equations of which the unknowns are part.
-
For example, the unknown 12x can be written as a product of the factors 12 and x. We can write 12x as 3 (4x), 2 (6x), etc., taking advantage of the factors of 12 that are more convenient for us.
We can also go further and break it down 12x more times. In other words, we don't have to stop at 3 (4x) or 2 (6x), but we can further break down 4x and 6x to get 3 (2 (2x) and 2 (3 (2x), respectively. Of course, these two expressions are equivalent
Step 3. Apply the distributive property to factor algebraic equations
By taking advantage of your knowledge of the decomposition of both single numbers and unknowns with coefficient, you can simplify basic algebraic equations by identifying factors common to both numbers and unknowns. Usually, to simplify the equations as much as possible, we try to find the greatest common divider. This simplification process is possible thanks to the distributive property of multiplication, which says that taking any numbers a, b, c, a (b + c) = ab + ac.
- Let's try an example. To break down the algebraic equation 12 x + 6, first of all we find the Greatest Common Divider of 12x and 6. 6 is the largest number that perfectly divides both 12x and 6, so we can simplify the equation into 6 (2x + 1).
- This procedure can also be applied to equations that contain negative numbers and fractions. x / 2 + 4, for example, can be simplified to 1/2 (x + 8), and -7x + -21 can be decomposed as -7 (x + 3).
Method 2 of 3: Factoring Second Degree (or Quadratic) Equations
Step 1. Make sure the equation is second degree (ax2 + bx + c = 0).
Second degree equations (also called quadratic) are in the form x2 + bx + c = 0, where a, b, and c are numeric constants and a is different from 0 (but it can be 1 or -1). If you find yourself with an equation that contains the unknown (x) and has one or more terms with x on the second member, you can move them all to the same member with basic algebraic operations to get 0 from one part of the equal sign and ax2, etc. on the other.
- For example, let's take the following algebraic equation. 5x2 + 7x - 9 = 4x2 + x - 18 can be simplified to x2 + 6x + 9 = 0, which is second degree.
- Equations with powers greater than x, such as x3, x4, etc. they are not second degree equations. These are equations of the third, fourth degree, and so on, unless the equation can be simplified by eliminating the terms with the x raised to a number greater than 2.
Step 2. In quadratic equations where a = 1, factor in (x + d) (x + e), where d × e = c and d + e = b
If the equation is of the form x2 + bx + c = 0 (that is, if the coefficient of x2 = 1), it is possible (but not certain) that a faster method could be used to break down the equation. Find two numbers that when multiplied together give c And added together give b. Once you find these numbers d and e, substitute them in the following formula: (x + d) (x + e). The two terms, when multiplied, result in the original equation; in other words, they are the factors of the quadratic equation.
- Take for example the second degree equation x2 + 5x + 6 = 0. 3 and 2 multiplied together give 6, while added together they give 5, so we can simplify the equation to (x + 3) (x + 2).
-
There are slight variations of this formula, based on some differences in the equation itself:
- If the quadratic equation is of the form x2-bx + c, the result will be like this: (x - _) (x - _).
- If it is in the form x2+ bx + c, the result will be like this: (x + _) (x + _).
- If it is in the form x2-bx-c, the result will be like this: (x + _) (x - _).
- Note: numbers in spaces can also be fractions or decimals. For example, the equation x2 + (21/2) x + 5 = 0 decomposes into (x + 10) (x + 1/2).
Step 3. If possible, break it down by trial and error
Believe it or not, for simple second-degree equations, one of the accepted methods of factoring is to simply examine the equation and then consider possible solutions until you find the right one. This is why it is called trial breaking. If the equation is of the form ax2+ bx + c and a> 1, the result will be written (dx +/- _) (ex +/- _), where d and e are numerical constants other than zero which multiplying give a. Both d and e (or both) can be the number 1, although not necessarily. If both are 1, you basically just used the quick method described earlier.
Let's proceed with an example. 3x2 - 8x + 4 at first glance can be intimidating, but just think that 3 has only two factors (3 and 1) and it will immediately seem simpler, since we know that the result will be written in the form (3x +/- _) (x +/- _). In this case, putting a -2 in both spaces will get the right answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x added to -8x. -2 × -2 = 4, so we can see that the factorized terms in brackets multiply to give the original equation.
Step 4. Solve by executing the square
In some cases, quadratic equations can be easily factored using a special algebraic identity. All second degree equations written in the form x2 + 2xh + h2 = (x + h)2. So, if in your equation the value of b is twice the square root of c, the equation can be factored into (x + (sqrt (c)))2.
For example, the equation x2 + 6x + 9 is suitable for demonstration purposes, because it is written in the right form. 32 is 9 and 3 × 2 is 6. We therefore know that the factorized equation will be written like this: (x + 3) (x + 3), or (x + 3)2.
Step 5. Use factors to solve second degree equations
Regardless of how you break down the quadratic expression, once you break it down you can find the possible values of x by setting each factor equal to 0 and solving. Since you have to figure out for which values of x the result is zero, the solution will be that one of the factors of the equation is equal to zero.
Let's go back to the equation x2 + 5x + 6 = 0. This equation breaks down to (x + 3) (x + 2) = 0. If one of the factors equals 0, the whole equation will also be equal to 0, so the possible solutions for x are the numbers that make (x + 3) and (x + 2) equal to 0. These numbers are -3 and -2, respectively.
Step 6. Check the solutions, as some may not be acceptable
When you have identified the possible values of x, substitute them one at a time in the starting equation to see if they are valid. Sometimes the found values, when substituted in the original equation, do not result in zero. These solutions are called "unacceptable" and must be discarded.
-
We substitute -2 and -3 in the equation x2 + 5x + 6 = 0. Before -2:
- (-2)2 + 5(-2) + 6 = 0
- 4 + -10 + 6 = 0
- 0 = 0. That's correct, so -2 is an acceptable solution.
-
Now let's try -3:
- (-3)2 + 5(-3) + 6 = 0
- 9 + -15 + 6 = 0
- 0 = 0. This result is also correct, so -3 is also an acceptable solution.
Method 3 of 3: Factoring Other Types of Equations
Factor Algebraic Equations Step 10 Step 1. If the equation is written in the form a2-b2, break it down into (a + b) (a-b).
Equations with two variables break down differently from normal second degree equations. For each equation a2-b2 with a and b different from 0, the equation breaks down into (a + b) (a-b).
For example, let's take the equation 9x2 - 4y2 = (3x + 2y) (3x - 2y).
Factor Algebraic Equations Step 11 Step 2. If the equation is written in the form a2+ 2ab + b2, break it down into (a + b)2.
Note that if the trinomial is written a2-2ab + b2, the factorized form is slightly different: (a-b)2.
The 4x equation2 + 8xy + 4y2 you can rewrite it as 4x2 + (2 × 2 × 2) xy + 4y2. Now we see that it is in the correct form, so we can say with certainty that it can be decomposed into (2x + 2y)2
Factor Algebraic Equations Step 12 Step 3. If the equation is written in the form a3-b3, break it down into (a-b) (a2+ ab + b2).
Finally, it must be said that the equations of third degree and beyond can also be factored, even if the procedure is significantly more complex.
For example, 8x3 - 27y3 breaks down into (2x - 3y) (4x2 + ((2x) (3y)) + 9y2)
Advice
- to2-b2 is decomposable, while a2+ b2 it is not.
- Remember how constants break down, it might be useful.
- Be careful when you have to work on the fractions, do all the steps carefully.
- If you have a trinomial written in the form x2+ bx + (b / 2)2, decomposed into (x + (b / 2))2 - you may find yourself in this situation when making a square.
- Remember that a0 = 0 (due to the multiplication by zero property).
