How to Factor a Quadratic Equation

In mathematics, factoring is finding numbers or expressions that have the product of a given number or equation. Factoring is a useful skill worth learning for solving basic algebra problems: the ability to factorize proficiently is almost the key when it comes to work. with algebraic equations or other forms of polynomials. Factoring can be used to reduce algebraic expressions, making math problems simpler. Thanks to it, you can even eliminate certain possible answers much faster than solving by hand.

Table of Contents

Factorize numbers and basic algebraic expressions

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Understand the definition of factoring as it applies to single numbers. Although conceptually simple, in practice applying complex equations can be quite challenging. Therefore, the easiest conceptual approach to factoring is to start with single numbers and then move on to simple equations before proceeding to more advanced applications. The factors of a given number are numbers whose product is equal to the number itself. For example, 1, 12, 2, 6, 3, and 4 are the factors of 12 because 1 × 12, 2 × 6, and 3 × 4 are all 12.

In other words, the factors of a given number are the numbers that are divisible by that number.
Can you find all the factors of 60? The number 60 is used for many different purposes (minutes in an hour, seconds in a minute, etc.)
- The number 60 has the following factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

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Understand that expressions containing variables can also be factored. Like independent numbers, variables with arithmetic coefficients can also be factored. To do so, we just need to find the factor of the coefficient of the variable. Knowing how to factor variables is useful for simply transforming algebraic equations that contain variables.

For example 12x can be rewritten as the product of 12 and x. We can write 12x as 3(4x), 2(6x), etc., and use whatever factor best suits the intended use of 12.
- You can even go as far as analyzing 12x multiple times . In other words, there’s no need to stop at 3(4x) or 2(6x) – we can break down 4x and 6x to get 3(2(2x) 2(3(2x) respectively). This formula is equivalent.

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Apply the associative property of multiplication to factor algebraic equations. Using your knowledge of factoring both independent numbers and variables, you can simplify algebraic equations simply by finding the common factors of the numbers and variables included in the equation. Usually, to keep the equation as simple as possible, we will try to find the greatest common divisor. This simple transformation is possible thanks to the associative property of multiplication – for all numbers a, b and c, we have: a(b + c) = ab + ac .

Let’s consider the following example problem. To factor the algebraic equation 12x + 6, we first find the greatest common divisor of 12x and 6. 6 is the largest number that both 12x and 6 are divisible by, so we can simply transform it. Simplify the equation to 6(2x + 1).
This process also applies to equations with negative signs and fractions. For example, x/2 + 4 can be simply transformed into 1/2(x + 8), and -7x + -21 can be decomposed into -7(x + 3).

Factoring quadratic equations

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Make sure the equation is in quadratic form (ax ² + bx + c = 0). The quadratic equation takes the form ax ² + bx + c = 0, where a, b, and c are constants and a is non-zero (note that a can be either 1 or -1). If the equation of one variable (x) contains one or more terms containing the square of x, you can usually use basic algebraic operations to transform, returning one side of the equal sign to zero and letting ax ² , etc on the other side.

For example, the algebraic equation 5x ² + 7x – 9 = 4x ² + x – 18 can be reduced to x ² + 6x + 9 = 0, which is the quadratic form.
Equations where x has a higher exponent, such as x ³ , x ⁴ , etc. cannot be a quadratic equation. They are quadratic, quaternary, etc. unless the equation can be reduced by suppressing terms containing powers of 3 or more of x.

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With a quadratic equation, when a = 1, we break it down as (x+d )(x+e), where d × e = c and d + e = b. If the quadratic equation is in the form x ² + bx + c = 0 (or in other words, if the coefficient of x ² = 1), it is possible (but unlikely) that we can use a quick calculation It is relatively simple to factorize this equation. Find two numbers whose product is c and sum is b. Once d and e are found, substitute them into the following expression: (x+d)(x+e) . When multiplied together, these two elements give the above quadratic equation – in other words, they are the factors of the equation.

Take for example the quadratic equation x ² + 5x + 6 = 0. 3 and 2 have a product of 6 and, at the same time, a sum of 5. Therefore, we can simply transform the equation into (x + 3) (x + 2).
This basic quick fix is a little different when the equation itself is a little different:
- If the quadratic equation was in the form x ² -bx+c, your answer would be: (x – _)(x – _).
- If it was in the form x ² +bx+c, your answer would be: (x + _)(x + _).
- If it was in the form x ² -bx-c, your answer would be in the form (x + _)(x – _).
Note: in the blanks can be fractions or decimals. For example, the equation x ² + (21/2)x + 5 = 0 is decomposed into (x + 10)(x + 1/2).

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If possible, perform the factorization by experiment. Believe it or not, with an uncomplicated quadratic equation, one of the accepted methods of factoring is simply to look at the problem, and then weigh the possible solutions until you find one. correct answer. It is also known as the test method. If the equation is of the form ax ² +bx+c and a>1, your factorization will be of the form (dx +/- _)(ex +/- _), where d and e are constants another number that does not have a product of a. d or e (or both) can equal 1, though not necessarily. If both equals 1, you’ve basically used the quick workaround described above.

Consider the following example problem. At first, 3x ² – 8x + 4 looks pretty scary. However, once we realize that 3 has only two factors (3 and 1), the problem becomes easier because we know the answer must be of the form (3x +/- _)(x +/- _). In this case, substituting -2 in both spaces will give the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x sum to -8x. -2 × -2 = 4, so it can be seen that the elements analyzed in brackets when multiplied together give the original equation.

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Solve the problem by completing the squaring. In some cases, quadratic equations can be factored quickly and easily using a special algebraic identity. Any quadratic equation of the form x ² + 2xh + h ² = (x + h) ² . Thus, if in the equation, b is twice the square root of c, the equation can be decomposed into (x + (sqrt(c)))) ² .

For example, the equation x ² + 6x + 9 would fit this form. 3 ² equals 9 and 3 × 2 equals 6. So we know that the factorized form of this equation is (x + 3)(x + 3), or (x + 3) ² .

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Solve quadratic equations by multiplying. Either way, once the quadratic expression has been factored, you can find a possible answer to the value of x by making each factor zero and solving. Since we are looking to find the value of x such that the equation is zero, any x that causes a factor to be zero will also be a possible solution of that equation.

Back to the equation x ² + 5x + 6 = 0. This equation is broken down to (x + 3)(x + 2) = 0. When one factor is zero, the whole equation will be zero. So , the possible solutions of x are the numbers that make (x + 3) and (x + 2) zero, -3 and -2 respectively.

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Check your answers – some may be outliers! When you find possible solutions for x, substitute them into the original equation to determine if they are correct. Sometimes, the solution found doesn’t make the original equation zero when substituted. We call these outliers and discard them.

Let’s replace -2 and -3 in x ² + 5x + 6 = 0. First, -2:
- (-2) ² + 5(-2) + 6 = 0
- 4 + -10 + 6 = 0
- 0 = 0. True, so -2 is a valid solution of the equation.
Now, try with -3:
- (-3) ² + 5(-3) + 6 = 0
- 9 + -15 + 6 = 0
- 0 = 0. It is also true and so -3 is also a valid solution of the equation.

Factoring other types of equations

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If the equation is in the form a ² -b ² , break it down as (a+b)(ab). A two-variable equation is analyzed differently than a basic quadratic equation. Any equation a ² -b ² in which a and b are non-zero will be decomposed as (a+b)(ab).

For example, the equation 9x ² – 4y ² = (3x + 2y)(3x – 2y).

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If the equation is in the form a ² +2ab+b ² , break it down to (a+b) ² . Note that if the trinomial is in the form a ²– 2ab+b ² , the factorization will be slightly different: (ab) ² .

The equation 4x ² + 8xy + 4y ² can be rewritten as 4x ² + (2 × 2 × 2)xy + 4y ² . We now see that it is in the correct form and can confidently say that the factorized form of this equation is (2x + 2y) ² .

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If the equation is in the form a ³ -b ³ , break it down as (ab)(a ² +ab+b ² ). Finally, it should be said that cubic equations and even higher order equations can be factored. However, the analysis process quickly becomes incredibly complicated.

For example, 8x ³ – 27y ³ is parsed as (2x – 3y)(4x ² + ((2x)(3y)) + 9y ² )

Advice

a ² -b ² can be factored, but a ² +b ² is not.
Remember how to factor the constant – it might help.
Pay attention to the fraction in the factoring process, handle it properly and appropriately.
For a triangle of the form x ² +bx+ (b/2) ² , its factorization form would be (x+(b/2)) ² (you’ll probably encounter this situation while completing the normalization). direction).
Remember that a0=0 (the property of multiplying by zero).

Things you need

Paper
Pencil
Math book (if needed)

Steps

Factorize numbers and basic algebraic expressions

Factoring quadratic equations

Factoring other types of equations

Advice

Things you need

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