## Trinomials

A trinomial is an algebraic expression that has three terms. An algebraic expression consists of variables and constants of one or more terms. These expressions use symbols or operations as separators such as +, –, ×, and ÷. A trinomial along with monomial, binomial, and polynomial are categorized under this algebraic expression. Let us learn more about trinomials, factoring trinomials, the formula for factoring trinomials along with solving a few examples.

## What is a Trinomial?

A trinomial is an algebraic expression that has three non-zero terms and has more than one variable in the expression. A trinomial is a type of polynomial but with three terms. A polynomial is an algebraic expression that has one or more terms and is written as

A polynomial can be referred by different names depending on the number of terms it has. The table below mentions the names.

Number of terms | Polynomial | Example |

1 | Monomial | xy |

2 | Binomial | x + y |

3 | Trinomial | x^{2} + xz + 1 |

## Perfect Square Trinomial

A perfect square trinomial is defined as an algebraic expression that is obtained by squaring a binomial expression. It is of the form ax^{2} + bx + c. Here a, b, and c are real numbers and a ≠ 0. For example, let us take a binomial (x + 2) and multiply it to (x + 2). The result obtained is x^{2} + 4x + 4. A perfect square trinomial can be decomposed into two binomials and the binomials when multiplied with each other gives the perfect square trinomial.

## Quadratic Trinomial

A quadratic trinomial is a type of algebraic expression with variables and constants. It is expressed in the form of ax^{2} + bx + c, where x is the variable and a, b, and c are non-zero constants and integers. The constant ‘a’ is known as a leading coefficient, ‘b’ is the linear coefficient, ‘c’ is the additive constant. A quadratic trinomial also describes the discriminant D where it defines the quantity of an expression and it is written as D = b^{2} – 4ac. The discriminant helps in classifying among the different cases of quadratic trinomials. If the value of a quadratic trinomial with a single variable is zero, then it is known as a quadratic equation i.e ax^{2} + bx + c = 0.

## How to Factor Trinomials?

Factoring a trinomial means expanding an equation into the product of two or more binomials. It is written as (x + m) (x + n). A trinomial can be factorized in many ways. Let’s discuss each case.

### Quadratic Trinomial in One Variable

The general form of quadratic trinomial formula in one variable is ax^{2} + bx + c, where a, b, c are constant terms and neither a, b, or c is zero. For the value of a, b, c, if b^{2} – 4ac > 0, then we can always factorize a quadratic trinomial. It means that ax^{2} + bx + c = a(x + h)(x + k), where h and k are real numbers. Now let’s learn how to factorize a quadratic trinomial with an example.

**Example: **Factorize: 3x^{2} – 4x – 4

**Solution:**

**Step 1:- **First multiply the coefficient of x^{2} and the constant term.

3 × -4 = -12

**Step 2:- **Break the middle term -4x such that on multiplying the resulting numbers, we get the result -12 (obtained from the first step).

-4x = -6x + 2x

-6 × 2 = -12

**Step 3:- **Rewrite the main equation by applying the change in the middle term.

3x^{2} – 4x – 4 = 3x^{2} – 6x + 2x – 4

**Step 4:- **Combine the first two terms and the last two terms, simplify the equation and take out any common numbers or expressions.

3x^{2} – 6x + 2x – 4 = 3x (x – 2) + 2(x – 2)

**Step 5:- **Again take (x – 2) common from both the terms.

3x (x – 2) + 2(x – 2) = (x – 2) (3x + 2)

Therefore, (x – 2) and (3x + 2) are the factors of 3x^{2} – 4x – 4.

### Quadratic Trinomial in Two Variable

There is no specific way to solve a quadratic trinomial in two variables. Let’s take an example.

**Example: **Factorize: x^{2} + 3xy + 2y^{2}

**Solution**

**Step 1: **These types of trinomials also follow the same rule as above, i.e., we need to break the middle term.

x^{2} + 3xy + 2y^{2 }= x^{2} + 2xy + xy + 2y^{2}

**Step 2: **Simplify the equation and take out common numbers of expressions.

x^{2} + 2xy + xy + 2y^{2 }= x (x + 2y) + y (x + 2y)

**Step 3: **Again take (x + 2y) common from both the terms.

x (x + 2y) + y (x + 2y) = (x + y) (x + 2y)

Therefore, (x + y) and (x + 2y) are the factors of x^{2} + 3xy + 2y^{2}

### If Trinomial is an Identity

Let’s see some algebraic identities that are mentioned in the table below:

Identity | Expanded Form |

(x + y)^{2} | x^{2} + 2xy + y^{2} |

(x – y)^{2} | x^{2} – 2xy + y^{2} |

(x^{2} – y^{2}) | (x + y) (x – y) |

**Example**: Factorize: 9x^{2} + 12xy + 4y^{2}

**Solution:**

**Step 1: **Identify which identity can be applied in the expression.

We can apply (x + y)^{2} = x^{2} + 2xy + y^{2}

**Step 2: **Rearrange the expression so that it can appear in the form of the above identity.

9x^{2} + 12xy + 4y^{2 }= (3x)^{2} + 2 × 3x × 2y + (2y)^{2}

**Step 3: **Once the expression is arranged in the form of the identity, write its factors.

(3x)^{2} + 2 × 3x × 2y + (2y)^{2 }= (3x + 2y)^{2} = (3x + 2y) (3x + 2y)

Therefore, (3x + 2y) is the factor of 9x^{2} + 12xy + 4y^{2}.

### Leading coefficient of 1

Let us look at an example.

**Example:** Factorize x^{2 }+ 7x + 12

Solution:

**Step 1:** Compare the given equation with the standard form to obtain the coefficients.

ax^{2} + bx + c is the standard form, comparing the equation x^{2} + 7x + 12 we get a = 1, b = 7, and c = 12

**Step 2:** Find the paired factors of c i.e 12 such that their sum is equal to b i.e 7.

The pair factor of 12 are (1, 12), (2, 6), and (3, 4). Therefore, the suitable pair is 3 and 4.

**Step 3: **Add each number to x separately.

(x + 3) (x + 4)

Therefore, (x + 3) (x + 4) are the factors for x^{2 }+ 7x + 12.

### Factorizing with GCF

When the trinomial needs to be factorized where the leading coefficient is not equal to 1, the concept of GCF(Greatest Common Factor) is applied. Let us see the steps:

- Write the trinomial in descending order, from highest to lowest power.
- Find the GCF by factorization.
- Find the product of the leading coefficient ‘a’ and the constant ‘c.’
- Find the factors of the product ‘a’ and ‘c’. Pick a pair that sums up to get the number instead of ‘b’.
- Rewrite the original equation by replacing the term “bx” with the chosen factors.
- Factor the equation by grouping.

## Factoring Trinomials Formula

A trinomial can be a perfect square or a non-perfect square. We have two formulas to factorize a perfect square trinomial. But for factorizing a non-perfect square trinomial, we do not have any specific formula, instead, we have a process.

- The factoring trinomials formulas of perfect square trinomials are:
a

^{2}+ 2ab + b^{2}= (a + b)^{2}a

^{2}– 2ab + b^{2}= (a – b)^{2}For applying either of these formulas, the trinomial should be one of the forms a

^{2}+ 2ab + b^{2}(or) a^{2}– 2ab + b^{2}. - The process of factoring a non-perfect trinomial ax
^{2}+ bx + c is:**Step 1:**Find ac and identify b.**Step 2:**Find two numbers whose product is ac and whose sum is b.**Step 3:**Split the middle term as the sum of two terms using the numbers from step – 2.**Step 4:**Factor by grouping.

To factorize a trinomial of the form ax^{2} + bx + c, we can use any of the below-mentioned formulas:

- a
^{2}+ 2ab + b^{2}= (a + b)^{2}= (a + b) (a + b) - a
^{2}– 2ab + b^{2}= (a – b)^{2}= (a – b) (a – b) - a
^{2}– b^{2}= (a + b) (a – b) - a
^{3}+ b^{3}= (a + b) (a^{2}– ab + b^{2}) - a
^{3}– b^{3}= (a – b) (a^{2}+ ab + b^{2})

## Examples on Trinomials

**Example 1:** Help Tim find the factors of x^{2} – 5x + 6.

**Solution:**

x^{2} – 5x + 6

= x^{2} – 3x – 2x + 3 × 2

= x(x – 3) – 2x + 6

= x(x – 3) – 2(x – 3)

= (x – 2)(x – 3)

Therefore, (x – 2)(x – 3) is the factors of x^{2} – 5x + 6.

**Example 2: **Help Clara find the factors of 15a^{2} + 38ab + 24b^{2}.

**Solution:**

Here we need to break the middle term.

First, multiply the coefficient of a^{2} and b^{2} = 15 × 24 = 360.

Now find two numbers such that on multiplication they give the result 360 and on addition, they give the result 38.

18 × 20 = 360 and 18 + 20 = 38

15a^{2} + 38ab + 24b^{2}

= 15a^{2} + 18ab + 20ab + 24b^{2}

= 3a(5a + 6b) + 4b(5a + 6b)

= (5a + 6b)(3a + 4b)

Therefore, (5a + 6b)(3a + 4b) is the factors for 15a^{2} + 38ab + 24b^{2}.

**Example 3:** If y – 3 is a factor of y^{2} + a – 6y, then find the value of a. Find the other factor of the trinomial.

**Solution:**

(y – 3) is a factor of y^{2} + a – 6y. Then if we put (y = 3) in the trinomial y^{2} + a – 6y, its value will be 0.

3^{2} + a – 6 × 3 = 0

9 + a – 18 = 0

a – 9 = 0

a = 9

Now factorize the trinomial y^{2} + a – 6y = y^{2} + 9 – 6y

The above trinomial is the expansion of the identity (x – y)^{2} = x^{2} – 2xy + y^{2}

y^{2} – 6y + 9

= y^{2 }– 2 × 3 × y + 3^{2} = (y – 3)^{2}

Therefore, a = 9 and y^{2} + a – 6y = (y – 3)^{2}.

*Example 1*

Factor 6x^{2 }+ x – 2

__Solution__

The GCF =1, therefore it is of no help.

Multiply the leading coefficient a and the constant c.

⟹ 6 * -2 = -12

List all factors of 12 and identify a pair that has a product of -12 and a sum of 1.

⟹ – 3 * 4

⟹ -3 + 4 = 1

Now, rewrite the original equation by replacing the term “bx” with the chosen factors

⟹ 6x^{2 }– 3x + 4x – 2

Factor the expression by grouping.

⟹ 3x (2x – 1) + 2(2x – 1)

⟹ (3x + 2) (2x – 1)

*Example 2*

Factor 2x^{2 }– 5x – 12.

__Solution__

2x^{2 }– 5x – 12

= 2x^{2 }+ 3x – 8x – 12

= x (2x + 3) – 4(2x + 3)

= (2x + 3) (x – 4)

*Example 3*

Factor 6x^{2 }-4x -16

__Solution__

The GCF of 6, 4 and 16 is 2.

Factor out the GCF.

6x^{2 }– 4x – 16 ⟹ 2(3x^{2 }– 2x – 8)

Multiply the leading coefficient “a” and the constant “c.”

⟹ 6 * -8 = – 24

Identify the paired factors of 24 with the sum of -2. In this case, 4 and -6 are the factors.

⟹ 4 + -6 = -2

Rewrite the equation by replacing the term “bx” with the chosen factors.

2(3x^{2 }– 2x – 8) ⟹ 2(3x^{2 }+ 4x – 6x – 8)

Factor by grouping and don’t forget to include the GCF in your final answer.

⟹ 2[x (3x + 4) – 2(3x + 4)]

⟹ 2[(x – 2) (3x + 4)]

*Example 4*

Factor 3x^{3} – 3x^{2} – 90x.

__Solution__

Since the GCF= 3x, factor it out;

3x^{3} – 3x^{2} – 90x ⟹3x (x^{2} – x – 30)

Find a pair of factors whose product is −30 and sum is −1.

⟹- 6 * 5 =-30

⟹ −6 + 5 = -1

Rewrite the equation by replacing the term “bx” with the chosen factors.

⟹ 3x [(x^{2} – 6x) + (5x – 30)]

Factor the equation;

⟹ 3x [(x (x – 6) + 5(x – 6)]

= 3x (x – 6) (x + 5)

*Example 5*

Factor 6z^{2} + 11z + 4.

__Solution__

6z^{2} + 11z + 4 ⟹ 6*z*^{2} + 3*z* + 8*z* + 4

⟹ (6*z*^{2} + 3*z*) + (8*z* + 4)

⟹ 3z (2z + 1) + 4(2z + 1)

= (2*z* + 1) (3*z* + 4)

## FAQs on Trinomials

### What is a Trinomial?

A trinomial is an algebraic expression that has three non-zero terms and has more than one variable in the expression. For example: x^{2} + 5y – 25, a^{3} – 16b + 10. These are trinomials as they have three terms i.e. coefficient, variables, and constants. A trinomial can have only one variable or two variables.

### What is a Perfect Square Trinomial?

Perfect square trinomials are the trinomials that follow the identity (a + b)^{2 }= a^{2} + 2ab +b^{2}. It is an algebraic expression that is obtained by squaring a binomial expression. It is of the form ax^{2} + bx + c. Here a, b, and c are real numbers, and a ≠ 0. For example, x^{2} + 2x + 1 and 4y^{2} −20y + 25.

### How Do You Factor a Trinomial?

A trinomial can be factored in the form x^{2} + bx + c. First, we need to find two integers (y and z) whose product sums up to c and through addition, it sums up to b. Once we find the two numbers, we rewrite the trinomial as x^{2} + yx + zx + c and use the grouping and distributive property of the factor to find out the factors of the expression. The factors will be (x + y) (x + z).

### What is a Quadratic Trinomial?

A quadratic trinomial is a polynomial with three terms and the degree of the trinomial must be 2. It means that the highest power of the variable cannot be greater than 2. For example: x^{2} + y^{2}+ xy and x^{2} +2x + 3xy. It does not mean that a quadratic trinomial always turns into a quadratic equation when we equate it to zero. A quadratic equation is a quadratic trinomial formula with only one variable.

### What is a Cubic Trinomial?

A cubic trinomial is a trinomial which has degree 3. For example, x^{3} +2x + 4 and 2y^{3}− 3y^{2} +1.

### What is the Formula to Factor a Trinomial?

The factoring trinomials formulas of perfect square trinomials are:

- a
^{2}+ 2ab + b^{2}= (a + b)^{2} - a
^{2}– 2ab + b^{2}= (a – b)^{2}

For applying either of these formulas, the trinomial should be one of the forms a^{2} + 2ab + b^{2} (or) a^{2} – 2ab + b^{2}.

**Factoring Trinomials**

**Learning Objective(s)**

· Factor trinomials with a leading coefficient of 1.

· Factor trinomials with a common factor.

· Factor trinomials with a leading coefficient other than 1.

**Introduction**

A **polynomial** with three terms is called a **trinomial**. Trinomials often (but not always!) have the form *x*^{2 }+ *bx* + *c*. At first glance, it may seem difficult to factor trinomials, but you can take advantage of some interesting mathematical patterns to factor even the most difficult-looking trinomials.

So, how do you get from 6*x*^{2} + 2*x* – 20 to (2*x* + 4)(3*x* −5)? Let’s take a look.

**Factoring Trinomials: x^{2} + bx + c**

Trinomials in the form *x*^{2} + *bx* + *c* can often be factored as the product of two **binomials**. Remember that a binomial is simply a two-term polynomial. Let’s start by reviewing what happens when two binomials, such as (*x* + 2) and (*x* + 5), are multiplied.

Factoring is the reverse of multiplying. So let’s go in reverse and factor the trinomial *x*^{2} + 7*x* + 10. The individual terms *x*^{2}, 7*x*, and 10 share no common factors. So look at rewriting *x*^{2} + 7*x* + 10 as *x*^{2} + 5*x* + 2*x* + 10.

And, you can group pairs of factors: (*x*^{2} + 5*x*) + (2*x* + 10)

Factor each pair: *x*(*x* + 5) + 2(*x* + 5)

Then factor out the common factor *x *+ 5: (*x* + 5)(*x* + 2)

Here is the same problem done in the form of an example:

How do you know how to rewrite the middle term? Unfortunately, you can’t rewrite it just any way. If you rewrite 7*x* as 6*x* + *x*, this method won’t work. Fortunately, there’s a rule for that.

Factoring Trinomials in the form To factor a trinomial in the form x^{2 }+ bx + cx^{2 }+ bx + c, find two integers, r and s, whose product is c and whose sum is b. Rewrite the trinomial as x^{2 }+ rx + sx + c and then use grouping and the distributive property to factor the polynomial. The resulting factors will be (x + r) and (x + s). |

For example, to factor *x*^{2} + 7*x* +10, you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10: 1 and 10, 2 and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite 7*x* as 2*x* + 5*x*, and continue factoring as in the example above. Note that you can also rewrite 7*x* as 5*x* + 2*x*. Both will work.

Let’s factor the trinomial *x*^{2} + 5*x* + 6. In this polynomial, the *b* part of the middle term is 5 and the *c* term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the *c* term, 6; on the right you’ll find the sums.

Factors whose product is 6 | Sum of the factors |

1 • 6 = 6 | 1 + 6 = 7 |

2 • 3 = 6 | 2 + 3 = 5 |

There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that 2 + 3 = 5. So 2*x* + 3*x* = 5*x*, giving us the correct middle term.

Note that if you wrote *x*^{2} + 5*x* + 6 as *x*^{2} + 3*x* + 2*x* + 6 and grouped the pairs as (*x*^{2} + 3*x*) + (2*x* + 6); then factored, *x*(*x* + 3) + 2(*x* + 3), and factored out *x* + 3, the answer would be (*x* + 3)(*x* + 2). Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.

Finally, let’s take a look at the trinomial *x*^{2} + *x* – 12. In this trinomial, the *c* term is −12. So look at all of the combinations of factors whose product is −12. Then see which of these combinations will give you the correct middle term, where *b* is 1.

Factors whose product is −12 | Sum of the factors |

1 • −12 = −12 | 1 + −12 = −11 |

2 • −6 = −12 | 2 + −6 = −4 |

3 • −4 = −12 | 3 + −4 = −1 |

4 • −3 = −12 | 4 + −3 = 1 |

6 • −2 = −12 | 6 + −2 = 4 |

12 • −1 = −12 | 12 + −1 = 11 |

There is only one combination where the product is −12 and the sum is 1, and that is when *r* = 4, and *s* = −3. Let’s use these to factor our original trinomial.

In the above example, you could also rewrite *x*^{2} + *x* – 12 as *x*^{2} – 3*x* + 4*x* – 12 first. Then factor *x*(*x* – 3) + 4(*x* – 3), and factor out (*x* – 3) getting (*x* – 3)(*x* + 4). Since multiplication is commutative, this is the same answer.

**Factoring Tips**

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.

While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

**Tips for Finding Values that Work**

When factoring a trinomial in the form *x*^{2 }+ *bx* + *c*, consider the following tips.

Look at the *c* term first.

o If the *c* term is a positive number, then the factors of *c* will both be positive or both be negative. In other words, *r* and *s* will have the same sign.

o If the *c* term is a negative number, then one factor of *c* will be positive, and one factor of *c* will be negative. Either *r* or *s* will be negative, but not both.

Look at the *b* term second.

o If the *c* term is positive and the *b* term is positive, then both *r* and *s* are positive.

o If the *c* term is positive and the *b* term is negative, then both *r *and *s* are negative.

o If the *c* term is negative and the *b* term is positive, then the factor that is positive will have the greater absolute value. That is, if *|r| > |s|, *then *r* is positive and *s *is negative.

o If the *c* term is negative and the *b* term is negative, then the factor that is negative will have the greater absolute value. That is, if *|r| > |s|, *then *r* is negative and *s *is positive.

After you have factored a number of trinomials in the form *x*^{2 }+ *bx* + *c*, you may notice that the numbers you identify for *r* and *s* end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.

Trinomial | x^{2} + 7x + 10 | x^{2} + 5x + 6 | x^{2} + x – 12 |

r and s values | r = + 5, s = + 2 | r = + 2, s = + 3 | r = + 4, s = –3 |

Factored form | (x + 5)(x + 2) | (x + 2)(x + 3) | (x + 4)(x – 3) |

Notice that in each of these examples, the *r* and *s* values are repeated in the factored form of the trinomial.

So what does this mean? It means that in trinomials of the form *x*^{2 }+ *bx* + *c* (where the coefficient in front of *x*^{2} is 1), if you can identify the correct *r* and *s* values, you can effectively skip the grouping steps and go right to the factored form. You may want to stick with the grouping method until you are comfortable factoring, but this is a neat shortcut to know about!

Jess is trying to use the grouping method to factor the trinomial *v*^{2} – 10*v* + 21. How should she rewrite the central *b* term, −10*v*?

A) +7*v* + 3*v*

B) −7*v* – 3*v*

C) −7*v* + 3*v*

D) +7*v* – 3*v*

Show

A) +7*v* + 3*v*

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers 7 and 3, 7 + 3 = +10, so this would provide the term 10*v* instead of −10*v*.) The correct answer is −7*v* – 3*v*.

B) −7*v* – 3*v*

Correct. Because the *c* term is positive and the *b* term is negative, both terms should be negative. Check: using the integers −7 and −3, −7 + −3 = −10 and −7 • −3 = 21, so this provides both terms −10*v* and 21 correctly.

C) −7*v* + 3*v*

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers −7 and 3, −7 + 3 = −4 and − 7 • 3 = −21, so this would provide −4v instead of −10*v* and −21 instead of 21.) The correct answer is

−7*v* – 3*v*.

D) +7*v* – 3*v*

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers 7 and −3, 7 + −3 = 4 and 7 • −3 = −21, so this would provide 4*v* instead of −10*v* and −21 instead of 21.) The correct answer is

−7*v* – 3*v*.

**Identifying Common Factors**

Not all trinomials look like *x*^{2} + 5*x* + 6, where the coefficient in front of the *x*^{2} term is 1. In these cases, your first step should be to look for common factors for the three terms.

Trinomial | Factor out Common Factor | Factored |

2x^{2} + 10x + 12 | 2(x^{2} + 5x + 6) | 2(x + 2)(x + 3) |

−5a^{2} − 15a − 10 | −5(a^{2} + 3a + 2) | −5(a + 2)(a + 1) |

c^{3} – 8c^{2} + 15c | c(c^{2} – 8c + 15) | c(c – 5)(c – 3) |

y^{4 }– 9y^{3} – 10y^{2} | y^{2}(y^{2 }– 9y – 10) | y^{2}(y – 10)(y + 1) |

Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.

**Factoring Trinomials: ax^{2} + bx + c**

The general form of trinomials with a leading coefficient of *a* is *ax*^{2} + *bx *+ *c*. Sometimes the factor of *a* can be factored as you saw above; this happens when *a* can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an *x*^{2} term, instead of an *ax*^{2} term.

However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.

Factoring Trinomials in the form To factor a trinomial in the form ax^{2} + bx + cax^{2 }+ bx + c, find two integers, r and s, whose sum is b and whose product is ac. Rewrite the trinomial as ax^{2}^{ }+ rx + sx + c and then use grouping and the distributive property to factor the polynomial. |

This is almost the same as factoring trinomials in the form *x*^{2} + *bx *+ *c*,as in this form* a *= 1.Now you are looking for two factors whose product is *a *• *c*, and whose sum is *b*.

Let’s see how this strategy works by factoring 6*z*^{2} + 11*z* + 4.

In this trinomial, *a* = 6, *b* = 11, and *c* = 4. According to the strategy, you need to find two factors, *r* and *s*, whose sum is *b* (11) and whose product is *ac* (or 6 • 4 = 24). You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since *ac* is positive and *b* is positive, you can be certain that the two factors you’re looking for are also positive numbers.)

Factors whose product is 24 | Sum of the factors |

1 • 24 = 24 | 1 + 24 = 25 |

2 • 12 = 24 | 2 + 12 = 14 |

3 • 8 = 24 | 3 + 8 = 11 |

4 • 6 = 24 | 4 + 6 = 10 |

There is only one combination where the product is 24 and the sum is 11, and that is when *r* = 3, and *s* = 8. Let’s use these values to factor the original trinomial.

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial 2*z*^{2} + 35*z* + 7, for instance. Can you think of two integers whose sum is *b* (35) and whose product is *ac* (2 · 7 = 14)? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.

Factor 3
A) (3
B) (3
C) (3
D) (3
Answer A) (3 Incorrect. The product of (3
B) (3 Correct. The product of (3
C) (3 Incorrect. The product of (3
D) (3 Incorrect. The product of (3 |

**Negative Terms**

In some situations, *a* is negative, as in −4*h*^{2} + 11*h* + 3. It often makes sense to factor out −1 as the first step in factoring, as doing so will change the sign of *ax*^{2} from negative to positive, making the remaining trinomial easier to factor.

**Example**

Note that the answer above can also be written as (−*h* + 3)(4*h* + 1) or (*h* – 3)( −4*h* – 1) if you multiply −1 times one of the other factors.

**Summary**

Trinomials in the form *x*^{2 }+ *bx* + *c* can be factored by finding two integers, *r* and *s*, whose sum is *b* and whose product is *c.* Rewrite the trinomial as *x*^{2 }+ *rx* + *sx* + *c *and then use grouping and the distributive property to factor the polynomial.

When a trinomial is in the form of* ax*^{2} + *bx* + *c*, where *a* is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form *ax*^{2} + *bx* + *c, *find two integers, *r* and *s*, whose sum is *b* and whose product is *ac.* Then rewrite the trinomial as *ax ^{2}*

^{ }+

*rx*+

*sx*+

*c*and use grouping and the distributive property to factor the polynomial.

When *ax*^{2} is negative, you can factor −1 out of the whole trinomial before continuing.

**How to Factor a Trinomial in 3 Easy Steps**

Learning how to factor a trinomial is an extremely important and useful algebra skill, but factoring trinomials can also be very tricky.

This free How to Factor a Trinomial step-by-step guide will teach you how to factor a trinomial when *a*=1 and when a does *a* does not equal one (more on what *a* refers to later) using a simple three-step process.

Before you learn how to factor a trinomial, lets do a quick review of some very important vocabulary and definitions related to trinomials.

**Trinomial Definition**

A** trinomial **is a polynomial that has three terms. The first time is an x^2 term, the second term is an x term, and the third term is a constant (just a number).

**Furthermore, when discussing trinomials, you will see references to vales for a, b, and c., where:**

*a = the x^2 term coefficient*

*b = the x term coefficient*

*c= the constant value*

For example:

**For this trinomial: a=1, b=5, and c=6.**

Note that, when a=1, the coefficient will not be written in front of the x^2 term.

**Factoring Trinomials: a = 1**

For the first few examples, lets learn how to factor a trinomial when a, the leading coefficient is 1.

**How to Factor a Trinomial Example #1**

For the first example, let’s factor the trinomial: x^2 + 6x + 8

Again, note that a=1 in this example.

Now let’s factor the trinomial:

**Step 1: Identify the values for b and c.**

In this example, b=6 and c=8.

**Step 2: Find two numbers that ADD to b and MULTIPLY to c.**

This step can take a little bit of trial-and-error.

For instance, you could pick 5 and 1 because 5+1=6. But 5 x 1 does not equal 8, so these numbers would not work.

However, if you chose 4 and 2, you can easily confirm that:

4 + 2 =6 (the value of b); and

4 x 2 = 8 (the value of c)

**Step 3: Use the numbers you picked to write out the factors and check**

For this example, the factors would be (x+2) and (x+4)

**Final Answer: (x+4)(x+2)**

You can check your answer by multiplying the two factors (binomials) together to see if the result is the original trinomial as follows:

**How to Factor a Trinomial Example #2**

Let’s get more practice factoring trinomials when a is 1.

Factor: x^2 + 9x + 20

**Step 1: Identify the values for b and c.**

In this example, b=9 and c=20.

**Step 2: Find two numbers that ADD to b and MULTIPLY to c.**

Finding the right numbers won’t always be as easy as it was in example 1.

To make factoring trinomials easier, write down all of the factors of c that you can think of.

In this case, c=20, so:

20 x 1 = 20

10 x 2 = 20

5 x 40 = 20

Remember that the two numbers have to multiply to c AND add to b.

**The only factors of 20 that meet both of these requirements are 4 and 5.**

**Step 3: Use the numbers you picked to write out the factors and check**

The last step is to write out the factors: (x+5)(x+4)

**How to Factor a Trinomial Example #3**

Factor: x^2 – 5x -24

Notice that this example includes subtraction signs, but the process of factoring is still the same.

**Step 1: Identify the values for b and c.**

In this example, b= -5 and c= -24.

**Step 2: Find two numbers that ADD to b and MULTIPLY to c.**

Again, you have to find two values to ADD to -5 and MULTIPLY to -24.

This step can get tricky when you’re dealing with negative numbers. Remember that a negative times a negative is positive, so only one of the values can be negative if the product has to be -24.

After writing out all of the possible factors, you can conclude that:

-8 x 3 = -24

and

-8 + 3 = -5

**Step 3: Use the numbers you picked to write out the factors and check**

The last step is to write out the factors: (x-8)(x+3)

**How to Factor a Trinomial Using the Completing the Square Formula**

Finally, there is an alternate method to factoring a trinomial that is called completing the square. This method applies to factoring quadratic equations (when a trinomial equals a value, namely zero).

You can learn more about factoring using the completing the square formula by checking our free step-by-step guide.

This **Complete Guide to the Completing the Square** includes several examples, a step-by-step tutorial, an animated video mini-lesson, and a free worksheet and answer key.

**Conclusion: How to Factor a Trinomial**

You can factor a trinomial of the form ax^2 + bx + c, when a=1, by using the following 3-step method:

**Step 1: Identify the values for b and c.**

**Step 2: Find two numbers that ADD to b and MULTIPLY to c.**

**Step 3: Use the numbers you picked to write out the factors and check**

## Common Factor

First check if there any common factors.

## Guess and Check

Maybe we can guess an answer?

Example: what are the factors of 2x^{2} + 7x + 3 ?

No common factors.

Let us try to **guess** an answer, and then check if we are right … we might get lucky!

That is not a very good method. So let us try something else.

## A Method For Simple Cases

Luckily there is a method that works in simple cases.

With the quadratic equation in this form:

Much better than guessing!

**Let’s see Steps 1 to 4 again, in one go**:

OK, let us try another example:

### Finding Those Numbers

The hardest part is finding two numbers that multiply to give ac, and add to give b.

Here is another example to help you:

## Why Factor?

Well, one of the big benefits of factoring is that we can find the **roots** of the quadratic equation (where the equation is zero).

All we need to do (after factoring) is find where each of the two factors becomes zero

## Graphing

We can also try graphing the quadratic equation. Seeing where it equals zero can give us clues

## The General Solution

There is also a general solution (useful when the above method fails), which uses the quadratic formula:

## Formula Sheet

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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