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Precalculus: Help and Review11 chapters | 88 lessons

Instructor:
*DaQuita Hester*

DaQuita has taught high school mathematics for six years and has a master's degree in secondary mathematics education.

Trinomials can have different leading coefficients and therefore, have to be factored differently. Learn how to factor and solve any trinomial. Then, test your knowledge with a quiz.

A **trinomial** is an equation that consists of three terms. For this lesson, we will examine trinomials written in the form *ax^2 + bx + c*, where *a*, the leading coefficient, does not equal zero.

To **factor** a trinomial means to rewrite it as a product of two binomials. This means that we are going to rewrite the trinomial in the form *(x + m) (x + n)*. Your task is to determine the value of *m* and *n*.

For factoring, trinomials are divided into two groups: those with a leading coefficient of 1 and those with a leading coefficient not equal to 1. Let's examine both.

Use the following steps to factor the trinomial *x^2 + 7x + 12*.

Step 1: **Determine the factor pairs of c that will add to get b**. For *x^2 + 7x + 12*, *a* = 1, *b* = 7, and *c* = 12. So to complete this step, we have to figure out which factor pairs of 12 will add together to equal 7. Let's list them. The factor pairs are 1 & 12, 2 & 6, and 3 & 4. The third pair is what we need, because the sum of these two numbers is 7, which is our *b*.

Step 2: **In separate parentheses, add each number to x**. Here, we are simply going to take our factor pair and add each one to

To ensure that we've factored correctly, let's multiply (x + 3) by (x + 4) to see if we get our original trinomial. When we distribute, we see that (x)(x) = x^2, (x)(4) = 4x, (3)(x) = 3x, and (3)(4) = 12. By combining our like terms, we get x^2 + 7x +12, which was our original trinomial. Therefore, we can conclude that our factoring was done correctly.

To factor trinomials with *a* not equal to 1, the process will be a little different. Let's walk through the steps below and use them to factor *2x^2 - 5x - 3*.

Step 1: **Multiply a and c together**. For this trinomial,

Step 2: **Identify the factor pairs of this product that will add together to equal b**. To complete this step, we must list the factor pairs of -6. They are -1 & 6, 1 & -6, 2 & -3, and -2 & 3. The pair that will add together to get

Step 3: **Re-write the original equation, but replace b with the correct factor pair**. Since -5 = 1 - 6, we will replace -5x with 1x - 6x. This gives us

Step 4: ** Group the equation into two parentheses, each with two terms. Then, factor each one**. Our two parentheses will be (2x^2 + x) and (-6x - 3). Let's factor them. From the first set of parentheses, we can factor out an *x* to get x*(2x + 1). In the second parenthesis, let's factor out -3. This leaves us with -3*(2x + 1). Now we can see that *2x^2 - 5x - 3* = (2x^2 + x)+(-6x - 3) = x*(2x + 1) + -3*(2x + 1). Notice that the equations left inside of both parentheses are the same. If we are factoring correctly, this should happen every time.

Step 5: ** Place the factored terms into a separate parenthesis**. The terms that we factored out of the parenthesis were *x* and *-3*. By placing them into their own parenthesis, we end up with (x - 3) and (2x + 1) as the factors of *2x^2 - 5x - 3*.

Let's see if we did this correctly and multiply (x - 3) by (2x + 1). From distributing, we see that (x)(2x) = 2x^2, (x)(1) = x, (-3)(2x) = -6x, and (-3)(1) = -3. After combining like terms, we see that the product is 2x^2 - 5x - 3, which is what we started with. We can now be sure that we have factored correctly.

Now that we are able to factor trinomials, let's practice solving them. To do so, we want *ax^2 + bx + c* to equal zero.

Take a look at the equation 3x^2 - 5x - 5 = -3.

Before we can factor, we must get our trinomial to equal zero. To achieve this, we should add 3 to both sides. This leaves us with 3x^2 - 5x - 2 = 0, and now we are ready to factor.

Our leading coefficient is not equal to 1, so let's make sure that we follow the appropriate steps from above and begin by multiplying *a* and *c*. From this, we get (3)(-2) = -6. Now it's time to list the factor pairs. They are -1 and 6, 1 and -6, 2 and -3, & -2 and 3. The pair that will add together to get - 5 is 1 and -6, and our equation becomes 3x^2 + x - 6x - 2 = 0.

After grouping, we get (3x^2 + x)(-6x - 2) = 0. Then, once we factor each set of parentheses, we have x(3x + 1) and -2(3x + 1). At this point, we give those factored terms a new sets of parentheses, and conclude that 3x^2 - 5x - 2 factored is (x - 2)(3x + 1).

By substitution, (3x^2 + x)(-6x - 2) = 0 becomes (x - 2)(3x + 1) = 0.

To have a product of zero, one of the parentheses in our equation must be equal to zero. Truth is, it could be either one of them. Therefore, let's set each parenthesis equal to zero and solve for x. For (x - 2) = 0, we have a solution of x = 2. For (3x + 1) = 0, we have a solution of -1/3.

We have solved the trinomial 3x^2 - 5x - 5 = -3 and determined that the two solutions are 2 and -1/3.

The steps followed to factor a trinomial are determined by whether the leading coefficient is equal to 1 or not equal to 1. To solve a trinomial by factoring, the trinomial must be equal to zero.

Once you are finished, you should be able to identify, factor, and solve a trinomial.

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Precalculus: Help and Review11 chapters | 88 lessons

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