solving quadratic equations by factoring examples with answers

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Solving Quadratic Equations Using Factoring

To solve an quadratic equation using factoring :

      1 . Transform the equation using standard form in which one side is zero.

      2 .  Factor the non-zero side.

      3 .  Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero).

      4 .  Solve each resulting equation.

Solve the equation, x 2 − 3 x − 10 = 0

Factor the left side: ( x − 5 ) ( x + 2 ) = 0

Set each factor to zero: x − 5 = 0     or     x + 2 = 0

Solve each equation: x = 5     or     x = − 2

The solution set is { 5 , − 2 } .

Solve the equation, 2 x 2 + 5 x = 12

Set the right side to zero: 2 x 2 + 5 x − 12 = 0

Factor the left side: ( 2 x − 3 ) ( x + 4 ) = 0

Set each factor to zero: 2 x − 3 = 0 or x + 4 = 0

Solve each equation: x = 3 2 or x = − 4

The solution set is { 3 2 , − 4 } .

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Course: Algebra 1   >   Unit 14

Solving quadratics by factoring.

• Quadratics by factoring (intro)
• Solving quadratics by factoring: leading coefficient ≠ 1
• Quadratics by factoring
• Solving quadratics using structure
• Solve equations using structure
• Quadratic equations word problem: triangle dimensions
• Quadratic equations word problem: box dimensions
• Solving quadratics by factoring review

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Video transcript

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators

• Factoring Quadratic Equations – Methods & Examples

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How to Factor a Quadratic Equation?

Factoring when the coefficient of x 2 is greater than 1, practice questions, factoring quadratic equations – methods & examples.

First of all, let’s take a quick review of the quadratic equation . A quadratic equation is a polynomial of a second degree, usually in the form of f(x) = ax 2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f (x).

Every quadratic equation has two values of the unknown variable, usually known as the roots of the equation (α, β). We can obtain the roots of a quadratic equation by factoring the equation.

For this reason, factorization is a fundamental step towards solving any equation in mathematics. Let’s find out.

Factoring a quadratic equation can be defined as the process of breaking the equation into the product of its factors. In other words, we can also say that factorization is the reverse of multiplying out.

To solve the quadratic equation ax 2  + bx + c = 0 by factorization, the following steps are used:

• Expand the expression and clear all fractions if necessary.
• Move all terms to the left-hand side of the equal to sign.
• Factorize the equation by breaking down the middle term.
• Equate each factor to zero and solve the linear equations

Solve: 2(x 2  + 1) = 5x

Expand the equation and move all the terms to the left of the equal sign.

⟹ 2x 2  – 5x + 2 = 0

⟹ 2x 2  – 4x – x + 2 = 0

⟹ 2x (x – 2) – 1(x – 2) = 0

⟹ (x – 2) (2x – 1) = 0

Equate each factor equal to zero and solve

⟹ x – 2 = 0 or 2x – 1 = 0

⟹ x = 2 or x = 1212

Therefore, the solutions are x = 2, 1/2.

Solve 3x 2 – 8x – 3 = 0

3x 2  – 9x + x – 3 = 0

⟹ 3x (x – 3) + 1(x – 3) = 0

⟹ (x – 3) (3x + 1) = 0

⟹ x = 3 or x = -13

Solve the following quadratic equation (2x – 3) 2  = 25

Expand the equation (2x – 3) 2  = 25 to get;

⟹ 4x 2 – 12x + 9 – 25 = 0

⟹ 4x 2  – 12x – 16 = 0

Divide each term by 4 to get;

⟹ x 2  – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

The are many methods of factorizing quadratic equations. In this article, our emphasis will be based on how to factor quadratic equations, in which the coefficient of x 2  is either 1 or greater than 1.

Therefore, we will use the trial and error method to get the right factors for the given quadratic equation.

Factoring when the Coefficient of x 2 is 1

To factorize a quadratic equation of the form x 2 + bx + c, the leading coefficient is 1. You need to identify two numbers whose product and sum are c and b, respectively.

CASE 1: When b and c are both positive

Solve the quadratic equation: x 2  + 7x + 10 = 0

List down the factors of 10:

1 × 10, 2 × 5

Identify two factors with a product of 10 and a sum of 7:

1 + 10 ≠ 7 2 + 5 = 7.

Verify the factors using the distributive property of multiplication.

(x + 2) (x + 5) = x 2  + 5x + 2x + 10 = x 2  + 7x + 10

The factors of the quadratic equation are:(x + 2) (x + 5)

Equating each factor to zero gives;

x + 2 = 0 ⟹x= -2

x + 5 = 0 ⟹ x = -5

Therefore, the solution is x = – 2, x = – 5

x 2 + 10x + 25.

Identify two factors with the product of 25 and sum of 10.

5 × 5 = 25, and 5 + 5 = 10

Verify the factors.

x 2 + 10x + 25 = x 2 + 5x + 5x + 25

= x (x + 5) + 5x + 25

= x (x + 5) + 5(x + 5)

= (x + 5) (x + 5)

Therefore, x = -5 is the answer.

CASE 2: When b is positive and c is negative

Solve x 2  + 4x – 5 = 0

Write the factors of -5.

1 × –5, –1 × 5

Identify the factors whose product is – 5 and sum is 4.

1 – 5 ≠ 4 –1 + 5 = 4

Verify the factors using the distributive property.

(x – 1) (x + 5) = x 2  + 5x – x – 5 = x 2  + 4x – 5 (x – 1) (x + 5) = 0

x – 1 = 0 ⇒ x = 1, or x + 5 = 0 ⇒ x = -5

Therefore, x = 1, x = -5 are the solutions.

CASE 3: When b and c are both negative

x 2  – 5x – 6

Write down the factors of – 6:

1 × –6, –1 × 6, 2 × –3, –2 × 3

Now identify factors whose product is -6 and sum is –5:

1 + (–6) = –5

Check the factors using the distributive property.

(x + 1) (x – 6) = x 2  – 6 x + x – 6 = x 2  – 5x – 6

Equate each factor to zero and solve to get; (x + 1) (x – 6) = 0

x + 1 = 0 ⇒ x = -1, or x – 6 = 0 ⇒ x = 6

Therefore, the solution is x=6, x = -1

CASE 4: When b is negative and c is positive

x 2  – 6x + 8 = 0

Write down all factors of 8.

–1 × – 8, –2 × –4

Identify factors whose product is 8 and sum is -6 –1 + (–8) ≠ –6 –2 + (–4) = –6

(x – 2) (x – 4) = x 2  – 4 x – 2x + 8 = x 2  – 6x + 8

Now equate each factor to zero and solve the expression to get;

(x – 2) (x – 4) = 0

x – 2 = 0 ⇒ x = 2, or x – 4 = 0 ⇒ x = 4

Factorize x 2 +8x+12.

Write down the factors of 12;

12 = 2 × 6 or = 4 × 3 Find factors whose sum is 8:

2 + 6 = 8 2 × 6 ≠ 8

Use distributive property to check the factors;

= x 2 + 6x +2x + 12 = (x 2 + 6x) +(2x + 12) = x(x+6) +2(x+6)

= x (x + 6) +2 (x + 6) = (x + 6) (x + 2)

Equate each factor to zero to get;

(x + 6) (x + 2)

Sometimes, the leading coefficient of a quadratic equation may be greater than 1. In this case, we can not solve the quadratic equation by the use of common factors.

Therefore, we need to consider the coefficient of x 2 and the factors of c to find numbers whose sum is b.

Solve 2x 2  – 14x + 20 = 0

Determine the common factors of the equation.

2x 2  – 14x + 20 ⇒ 2(x 2  – 7x + 10)

Now we can find the factors of (x 2  – 7x + 10). Therefore, write down factors of 10:

–1 × –10, –2 × –5

Identify factors whose sum is – 7:

1 + (–10) ≠ –7 –2 + (–5) = –7

Check the factors by applying distributive property.

2(x – 2) (x – 5) = 2(x 2  – 5 x – 2x + 10) = 2(x 2  – 7x + 10) = 2x 2  – 14x + 20

Equate each factor to zero and solve; 2(x – 2) (x – 5) = 0

x – 2 = 0 ⇒ x = 2, or x – 5 = 0 ⇒ x = 5

Solve 7x 2  + 18x + 11 = 0

Write down the factors of both 7 and 11.

11 = 1 × 11

Apply distributive property to check the factors as shown below:

(7x + 1) (x + 11) ≠ 7x 2  + 18x + 11

(7x + 11) (x + 1) = 7x 2  + 7x + 11x + 11 = 7x 2  + 18x + 11

Now equate each factor to zero and solve to get;

7x 2  + 18x + 11= 0 (7x + 11) (x + 1) = 0

x = -1, -11/7

Solve 2x 2  − 7x + 6 = 3

2x 2  − 7x + 3 = 0

(2x − 1) (x − 3) = 0

x=1/2​ or x=3

Solve 9x 2 +6x+1=0

Factorize to give:

(3x + 1) (3x + 1) = 0

(3x + 1) = 0,

Therefore, x = −1​/3

Factorize 6x 2 – 7x + 2 = 0

6x 2  – 4x – 3x + 2 = 0

Factorize the expression;

⟹ 2x (3x – 2) – 1(3x – 2) = 0

⟹ (3x – 2) (2x – 1) = 0

⟹ 3x – 2 = 0 or 2x – 1 = 0

⟹ 3x = 2 or 2x = 1

⟹ x = 2/3 or x = ½

Factorize x 2  + (4 – 3y) x – 12y = 0

Expand the equation;

x 2  + 4x – 3xy – 12y = 0

⟹ x (x + 4) – 3y (x + 4) = 0

x + 4) (x – 3y) = 0

⟹ x + 4 = 0 or x – 3y = 0

⟹ x = -4 or x = 3y

Thus, x = -4 or x = 3y

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Factoring Quadratics

Factoring quadratics is a method of expressing the polynomial as a product of its linear factors. It is a process that allows us to simplify quadratic expressions, find their roots and solve equations. A quadratic polynomial is of the form ax 2 + bx + c, where a, b, c are real numbers. Factoring quadratics is a method that helps us to find the zeros of the quadratic equation ax 2 + bx + c = 0.

In this mini-lesson, let us learn about the fascinating concept of factoring quadratics, the formula for factorization of quadratic equations along some solved examples for a better understanding.

What Is Factoring Quadratics?

Factoring quadratics is a method of expressing the quadratic equation ax 2 + bx + c = 0 as a product of its linear factors as (x - k)(x - h), where h, k are the roots of the quadratic equation ax 2 + bx + c = 0. This method is also is called the method of factorization of quadratic equations. Factorization of quadratic equations can be done using different methods such as splitting the middle term, using the quadratic formula, completing the squares, etc.

Factoring Quadratics Meaning

The factor theorem relates the linear factors and the zeros of any polynomial. Every quadratic equation has two roots, say \(\alpha\) and \(\beta\). They are the zeros of the quadratic equation. Consider a quadratic equation f(x) = 0, where f(x) is a polynomial of degree 2. Suppose that x = \(\alpha\) is one root of this equation. This means that x = \(\alpha\) is a zero of the quadratic expression f(x). Thus, (x - \(\alpha\)) should be a factor of f(x).

Similarly, if x = \(\beta\) is the second root of f(x) = 0, then x = \(\beta\) is a zero of f(x). Thus, (x - \(\beta\)) should be a factor of f(x). Hence, factoring quadratics is a method of expressing the quadratic equations as a product of its linear factors, that is, f(x) = (x - \(\alpha\))(x - \(\beta\)). Let us go through some examples of factoring quadratics:

Examples of Factoring Quadratics

1. Consider the quadratic equation x 2 + 5x + 6 = 0

-3 and -2 are the roots of the equation. Verify by substituting the roots in the given equation and check if the value equals 0.

Factor 1: (x + 3)

LHS = x 2 + 5x + 6 = (-3) 2 + 5 × -3 + 6 = 9 -15 + 6 = 0 = RHS

Factor 2: (x + 2)

LHS = x 2 + 5x + 6 = (-2) 2 + 5 × -2 + 6 = 4 -10 + 6 = 0 = RHS

Thus the equation has 2 factors (x + 3) and (x + 2)

2. Consider x 2 - 9 = 0

3 and -3 are the two roots of the equation. Verify by substituting the roots in the given equation and check if the value equals 0.

3 2 - 9 = 9 - 9 = 0

(-3) 2 - 9 = 9 - 9 = 0

Thus the equation has 2 factors (x+3) and (x-3)

Methods of Factoring Quadratics

Factoring quadratics gives us the roots of the quadratic equation. There are different methods that can be used for factoring quadratic equations. Factoring quadratics is done in 4 ways:

• Factoring out the GCD
• Splitting the middle term
• Using Algebraic Identities (Completing the Squares)
• Using Quadratic formula

Factoring Quadratics by Taking Out The GCD

Factoring quadratics can be done by finding the common numeric factor and the algebraic factors shared by the terms in the quadratic equation and then take them out. Let us solve an example to understand the factoring quadratic equations by taking the GCD out.

Consider this quadratic equation: 3x 2 + 6x = 0

• The numerical factor is 3 (coefficient of x 2 ) in both terms.
• The algebraic common factor is x in both terms.
• The common factors are 3 and x. Hence we take them out.
• Thus 3x 2 + 6x = 0 is factorized as 3x(x + 2) = 0

Splitting the Middle Term for Factoring Quadratics

• The sum of the roots of the quadratic equation ax 2 + bx + c = 0 is given by \(\alpha + \beta\) = -b/a
• The product of the roots in the quadratic equation ax 2 + bx + c = 0 is given by \(\alpha\beta\) = c/a

We split the middle term b of the quadratic equation ax 2 + bx + c = 0 when we try to factorize quadratic equations. We determine the factor pairs of the product of a and c such that their sum is equal to b.

Example: f(x) = x 2 + 8x + 12

Split the middle term 8x in such a way that the factors of the product of 1 and 12 add up to make 8. Factor pairs of 12 are (1, 12), (2, 6), (3, 4). Now, we can see that the factor pair (2, 6) satisfies our purpose as the sum of 6 and 2 is 8 and the product is 12. Hence, we split the middle term and write the quadratic equation as:

x 2 + 8x + 12 = 0

⇒ x 2 + 6x + 2x + 12 = 0

Now, club the terms in pairs as:

(x 2 + 6x) + (2x + 12) = 0

⇒ x(x + 6) + 2(x + 6) = 0

Taking the common factor (x + 6) out, we have

(x + 2) (x + 6) = 0

Thus, (x + 2) and (x + 6) are the factors of x 2 + 8x + 12 = 0

Identities for Factoring Quadratics

The process of factoring quadratics can be done by completing the squares which require the use of algebraic identities. The main algebraic identities which are used for completing the squares are:

• (a + b) 2 = a 2 + 2ab + b 2
• (a - b) 2 = a 2 - 2ab + b 2

Steps to factorize quadratic equation ax 2 + bx + c = 0 using completeing the squares method are:

• Step 1: Divide both the sides of quadratic equation ax 2 + bx + c = 0 by a. Now, the obtained equation is x 2 + (b/a) x + c/a = 0
• Step 2: Subtract c/a from both the sides of quadratic equation x 2 + (b/a) x + c/a = 0. Obtained equation is x 2 + (b/a) x = -c/a
• Step 3: Add the square of (b/2a) to both the sides of quadratic equation x 2 + (b/a) x = -c/a. Obtained equation is x 2 + (b/a) x + (b/2a) 2 = -c/a + (b/2a) 2
• Step 4: Now the LHS of the quadratic equation x 2 + (b/a) x + (b/2a) 2 = -c/a + (b/2a) 2 can be written as a complete square and simplify the RHS, if necessary. Obtained equation is (x + b/2a) 2 = -c/a + (b/2a) 2
• Step 5: The roots of the given quadratic equation can be obtained and hence, we can form the factors of the equation.

Another algebraic identity which is used for factoring quadratics is a 2 - b 2 = (a + b)(a - b). Let us see an example to understand.

Example 1: f(x) = 9x 2 - 4 (difference of 2 perfect squares)

9x 2 - 4 = (3x) 2 - 2 2

We notice that this is of the form, a 2 - b 2 = (a + b)(a - b)

Hence we factorize the equation 9x 2 - 4 = 0 as (3x+2) (3x-2)

9x 2 - 4 = (3x+2) (3x-2)

Example 2: f(x) = 4x 2 + 12x + 9

4x 2 + 12x + 9 = (2x) 2 + 2(2x)(3) + (3) 2 We notice that this is of the form (a+b) 2 = a 2 + 2ab + b 2 (2x) 2 + 2(2x)(3) + (3) 2 = (2x + 3) 2 Hence we have (2x + 3), (2x + 3) as the linear factors of f(x) = 4x 2 + 12x + 9

Formula for Factoring Quadratics

Factoring quadratics is also done by using a formula that gives us the roots of the quadratic equation and hence, the factors of the equation. If ax 2 + bx + c = 0 is a quadratic equation, a is the coefficient of x 2 , b is the coefficient of x and c is the constant term. Then we find the value of x by using the formula:

\(x= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Consider, the quadratic equation x 2 + 5x + 4 = 0. It is of the form ax 2 + bx + c = 0. Here a =1, b = 5, c = 4

Substituting the values of a, b and c in the quadratic formula formula, we get

\[\begin{align}x&= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x &= \dfrac{-5\pm \sqrt{5^2-4\times 1 \times 4}}{2\times 1}\\\\&= \dfrac{-5\pm \sqrt{25 -16}}{2}\\\\&= \dfrac{-5\pm \sqrt{9}}{2}\\\\&= \dfrac{-5\pm 3}{2}\\&= \dfrac{-5+3}{2}\text{ and } \dfrac{-5-3}{2}\\\\&=\dfrac{-2}{2}\text{ and }\dfrac{-8}{2}\\\\x &=(-1)\text{ and} (-4)\end{align}\]

Thus, the factors are (x + 1) and (x + 4).

Related Topics on Factoring Quadratics

• Factor Theorem
• Factoring Methods

Important Notes on Factoring Quadratics

• Linear factors are of the form ax + b and they cannot be factored further.
• A quadratic polynomial is a polynomial of degree 2.

Factoring Quadratics Examples

Example 1: Verify if (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x 2 + 9x + 9.

Solution: To verify if (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x 2 + 9x + 9, we will multiply the factors.

(2x + 3)(x + 3) = 2x 2 + 3x + 6x + 9 = 2x 2 + 9x + 9.

Answer: Hence, (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x 2 + 9x + 9.

Example 2: Find the factors of the quadratic equation x 2 + x - 12 = 0 using the factoring quadratics method.

Solution: We will split the middle term of the quadratic equation x 2 + x - 12 = 0 to determine its factors.

x 2 + x - 12 = 0

⇒ x 2 + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

Answer: Hence the factors of x 2 + x - 12 = 0 are (x - 3) and (x + 4).

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Practice Questions on Factoring Quadratics

Faqs on factoring quadratics, what is factoring quadratics in algebra.

Factoring quadratics is a method of expressing the polynomial as a product of its linear factors. It is a process that allows us to simplify quadratic expressions, find their roots and solve equations.

How to Solve Quadratic Equations by Factoring Quadratics?

By determining the factors, we can get the roots of the quadratic equation and hence the solution.

What are the Methods used for Factorization of Quadratic Equations?

The methods to factorize quadratic equations are splitting the middle term, using algebraic identities, using the quadratic formula, and factoring the GCD out.

What is the Method of Factoring Quadratic Equations?

Splitting the middle term and using the quadratic formula are the most efficient methods for factoring quadratic equations.

Is Factoring Quadratics the Same as Solving it?

When we factorize a quadratic equation , we get linear factors that divide the quadratic polynomial evenly. The next step is finding the zeros of the equation by equating the factors with zero.

What is the Shortest Way to Factor any Quadratic Equation?

Using the quadratic formula is the shortest way of factoring quadratics.

What Are Some Tricks for Factoring Quadratics?

Find the sum of the roots and the product of the roots or by identifying any known algebraic identity, we can factorize the quadratic equations.

What is Quadratic Equation Factored Form?

(x - \(\alpha\)) (x - \(\beta\)) is the factored form of quadratic equation, where \(\alpha\) and \(\beta\) are the roots of the quadratic equation.

How to Solve a Quadratic Equation?

There are different methods that can be used for factoring quadratic equations and solving the quadratic equations. Factoring quadratics is done in 4 ways:

How Can you Factorize Quadratic Equations Easily?

We split the middle term b of the quadratic equation ax 2 + bx + c = 0 when we try to factorize quadratic equations. We determine the factor pairs of the product of a and c such that their sum is equal to b. Taking the common factors out, we can factorize quadratic equations easily.

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Solving Quadratic Equations by Factoring

Factoring Roots Completing the Square Formula Graphing Examples

This lesson covers many ways to solve quadratics, such as taking square roots, completing the square, and using the Quadratic Formula . But we'll start with solving by factoring.

(Before reaching the topic of solving quadratic equations, you should already know how to factor quadratic expressions. If not, first review how to factor quadratics .)

Content Continues Below

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You've already factored quadratic expressions. The new thing here is that the quadratic expression is part of an equation, and you're told to solve for the values of the variable that make the equation true.

How to solve a quadratic equation by factoring

• Put the quadratic expression on one side of the "equals" sign, with zero on the other side.
• Factor the quadratic expression into its two linear factors.
• Set each of these linear factors equal to zero, creating two linear equations.
• Solve the two linear equations.

Note: This process finds the zeroes (or solutions, or roots, or x -intercepts) of the quadratic by using the fact that a multiplication is equal to zero only if at least one of the factors being multiplied is equal to zero. (This is called the "Zero-Product Property".) This property says something that seems fairly obvious, but only after it's been pointed out to us; namely:

Zero-Product Property : If we multiply two (or more) things together and the result is equal to zero, then we know that at least one of those things that we multiplied must also have been equal to zero. Put another way, the only way for us to get zero when we multiply two (or more) factors together is for one of the factors to have been zero.

So, if we multiply two (or more) factors and get a zero result, then we know that at least one of the factors was itself equal to zero. In particular, we can set each of the factors equal to zero, and solve the resulting equation for one solution of the original equation.

We can only draw the helpful conclusion about the factors of the quadratic (namely, that one of those factors must have been equal to zero, so we can set the factors equal to zero) if the product itself equals zero. If the product of factors is equal to anything non-zero, then we can not make any claim about the values of the factors.

Therefore, when solving quadratic equations by factoring, we must always have the equation in the form "(quadratic expression) equals (zero)" before we make any attempt to solve the quadratic equation by factoring.

• Solve ( x − 3)( x − 4) = 0 by factoring.

Okay, this quadratic is already factored for me. But how do I use this factorisation to solve the equation?

The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, then I can set each of the factors equal to zero:

x − 3 = 0   or   x − 4 = 0

This gives me simple linear equations, and they're easy to solve :

x = 3   or   x = 4

And these two values are the solution they're looking for:

Note that " x = 3, 4 " means the same thing as " x  = 3 or x  = 4 "; the only difference is the formatting. The " x  = 3, 4 " format is more common.

• Solve x 2  + 5 x  + 6 = 0 , and check.

This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example, this isn't yet factored. I MUST factor the quadratic first, because it is only when I MULTIPLY and get zero that I can say anything about the factors and solutions. I can't conclude anything about the individual terms of the unfactored quadratic (like the 5 x or the 6 ), because I can add lots of stuff that totals to zero.

So the first thing I have to do is factor:

x 2 + 5 x + 6 = ( x + 2)( x + 3)

Now I can restate the original equation in terms of a product of factors, with this product being equal to zero:

( x + 2)( x + 3) = 0

Now I can solve each factor by setting each one equal to zero and solving the resulting linear equations:

x + 2 = 0   or   x + 3 = 0

x = −2   or   x = − 3

These two values are the solution to the original quadratic equation. So my answer is:

x = −3, −2

I'm not done, though, because the original exercise told me to "check", which means that I need to plug my answers back into the original equation, and make sure it comes out right. In this case, I'll be plugging into the expression on the left-hand side of the original equation, and verifying that I end up with the right-hand side; in other words, I need to confirm that plugging in these solution values results with the left-hand side equalling 0 :

checking x = −3 :

[−3] 2 + 5[−3] + 6

9 − 15 + 6

9 + 6 − 15

15 − 15

checking x = −2 :

[−2] 2 + 5[−2] + 6

4 − 10 + 6

4 + 6 − 10

10 − 10

When an exercise specifies that you should solve "and check", they're looking for you to show that you've plugged your answers into the original exercise and gotten something that worked out right. The above, where I showed my checks, is all they're wanting. Just make sure to do your work neatly.

By the way, you can use this "checking" technique to verify your answers to any "solving" exercise. So, for instance, if you're not sure of your answer to a "factor and solve" question on the next test, try plugging your answers into the original equation, and confirming that your solutions lead to true statements.

• Solve x 2 − 3 = 2 x .

This equation is not in "(quadratic) equals (zero)" form, so I can't try to solve it yet. The first thing I need to do is get all the terms over on one side, with zero on the other side. Only then can I factor and solve:

x 2 − 3 = 2 x

x 2 − 2 x − 3 = 0

( x − 3)( x + 1) = 0

x − 3 = 0, x + 1 = 0

x = 3, x = −1

Then my solution is:

x = −1, 3

• Solve ( x + 2)( x + 3) = 12 .

It is very common for students to see this type of equation, and say:

"Cool! It's already factored! So I'll set the factors equal to 12 and solve to get x = 10 and x = 9. That was easy!"

Yeah, that was easy; it was also wrong. Very, very wrong.

Besides the fact that neither (10 + 2)(10 + 3) nor (9 + 2)(9 + 3) equals 12 , we should never forget that we must have "(quadratic) equals (zero)" before we can solve by factoring.

Returning to the exercise:

Tempting though it may be, I cannot set each of the factors on the left-hand side of the equation equal to the other side of the equation and solve. Doing so would give me an entirely-wrong mess.

Instead, I first have to multiply out and simplify the left-hand side, then subtract the 12 over to the left-hand side, and re-factor. Only then can I solve.

( x + 2)( x + 3) = 12

x 2 + 5 x + 6 = 12

x 2 + 5 x − 6 = 0

( x + 6)( x − 1) = 0

x + 6 = 0, x − 1 = 0

x = −6, x = 1

x = −6, 1

• Solve x 2 + 5 x = 0 .

This two-term quadratic is easier to factor than were the previous quadratics: I see immediately that I can factor an x out of both terms, taking the x out front. This gives me:

x ( x + 5) = 0

A very common mistake that students make at this stage is to "solve" the equation for " x  + 5 = 0 " by dividing through by the x . But that's an invalid step. Why? Because we can't divide by zero. How does that come into play here?

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Dividing through by the factor x makes the implicit assumption that x was not equal to zero. There is absolutely no justification for making that assumption! And making that assumption would cause us to lose half of our solution to this equation.

I need to remember that it's okay for a factor to contain only a variable, without being added to other terms; in particular, " x " is a perfectly valid factor. I need to set both of the factors equal to zero, and then solve the two resulting linear equations:

x = 0, x + 5 = 0

x = 0, x = −5

x = 0, −5

The previous example had two terms and was easy to factor. There is one other case of two-term quadratics that we can factor to solve. It's only a bit more complicated:

• Solve x 2 − 4 = 0 .

This equation is in "(quadratic) equals (zero)" form, so it's ready for me to solve by factoring. But how do I factor this? By noticing that this is a difference of squares . I'll apply the difference-of-squares formula that I've memorized:

x 2 − 4 = 0

( x − 2)( x + 2) = 0

x − 2 = 0, x + 2 = 0

x = 2, x = −2

x = −2, 2

Note: The solution above can also be formatted as " x = ± 2" . This is pronounced as " x is equal to plus or minus 2 ".

The last example above leads us into how to solve by taking square roots, on the next page.

You can use the Mathway widget below to practice solving quadratic equations by factoring. Try the entered exercise, or type in your own exercise. Then click the button and select "Solve by factoring" to compare your answer to Mathway's. (Or skip ahead to the next page.)

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URL: https://www.purplemath.com/modules/solvquad.htm

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