A polynomial with three terms is called a trinomial. Trinomials often (but not always!) have the form x2 + 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 6x2 + 2x – 20 to (2x + 4)(3x −5)? Let’s take a look.
Trinomials in the form x2 + 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.
Example
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Problem
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Multiply (x + 2)(x + 5).
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(x + 2)(x + 5)
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Use the FOIL method to multiply binomials.
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x2 + 5x + 2x +10
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Then combine like terms 2xand 5x.
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Answer
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x2 + 7x +10
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Factoring is the reverse of multiplying. So let’s go in reverse and factor the trinomial x2 + 7x + 10. The individual terms x2, 7x, and 10 share no common factors. So look at rewriting x2 + 7x + 10 as x2 + 5x + 2x + 10.
And, you can group pairs of factors: (x2 + 5x) + (2x + 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:
Example
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Problem
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Factor x2 + 7x +10.
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x2 + 5x + 2x +10
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Rewrite the middle term 7x as 5x + 2x.
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x(x + 5) + 2(x + 5)
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Group the pairs and factor out the common factor x from the first pair and 2 from the second pair.
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(x + 5)(x + 2)
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Factor out the common factor
(x + 5).
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Answer
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(x + 5)(x + 2)
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How do you know how to rewrite the middle term? Unfortunately, you can’t rewrite it just any way. If you rewrite 7x as 6x + x, this method won’t work. Fortunately, there's a rule for that.
Factoring Trinomials in the form x2 + bx + c
To factor a trinomial in the form x2 + bx + c, find two integers, r and s, whose product is c and whose sum is b.
Rewrite the trinomial as x2 + 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).
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For example, to factor x2 + 7x +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 7x as 2x + 5x, and continue factoring as in the example above. Note that you can also rewrite 7x as 5x + 2x. Both will work.
Let’s factor the trinomial x2 + 5x + 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 cterm, 6; on the right you'll find the sums.
Factors whose product is 6
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Sum of the factors
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1 • 6 = 6
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1 + 6 = 7
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2 • 3 = 6
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2 + 3 = 5
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There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that 2 + 3 = 5. So 2x + 3x = 5x, giving us the correct middle term.
Example
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Problem
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Factor x2 + 5x + 6.
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x2 + 2x + 3x + 6
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Use values from the chart above. Replace 5x with 2x + 3x.
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(x2 + 2x) + (3x + 6)
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Group the pairs of terms.
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x(x + 2) + (3x + 6)
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Factor x out of the first pair of terms.
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x(x + 2) + 3(x + 2)
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Factor 3 out of the second pair of terms.
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(x + 2)(x + 3)
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Factor out (x + 2).
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Answer
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(x + 2)(x + 3)
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Note that if you wrote x2 + 5x + 6 as x2 + 3x + 2x + 6 and grouped the pairs as (x2 + 3x) + (2x + 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 x2 + 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
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Sum of the factors
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1 • −12 = −12
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1 + −12 = −11
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2 • −6 = −12
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2 + −6 = −4
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3 • −4 = −12
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3 + −4 = −1
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4 • −3 = −12
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4 + −3 = 1
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6 • −2 = −12
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6 + −2 = 4
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12 • −1 = −12
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12 + −1 = 11
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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.
Example
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Problem
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Factor x2 + x – 12
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x2 + 4x + −3x – 12
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Rewrite the trinomial using the values from the chart above. Use values r = 4 and s = −3.
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(x2 + 4x) + (−3x – 12)
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Group pairs of terms.
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x(x + 4) + (−3x – 12)
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Factor x out of the first group.
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x(x + 4) – 3(x + 4)
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Factor −3 out of the second group.
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(x + 4)(x – 3)
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Factor out (x + 4).
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Answer
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(x + 4)(x – 3)
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In the above example, you could also rewrite x2 + x – 12 as x2 – 3x + 4x – 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 x2 + 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.
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After you have factored a number of trinomials in the form x2 + 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
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x2 + 7x + 10
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x2 + 5x + 6
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x2 + x - 12
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r and s values
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r = + 5, s = + 2
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r = + 2, s = + 3
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r = + 4, s = –3
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Factored form
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(x + 5)(x + 2)
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(x + 2)(x + 3)
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(x + 4)(x – 3)
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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 x2 + bx + c (where the coefficient in front of x2 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!
Source: https://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U12_L2_T1_text_final.html
Algebra 2
Here are the steps required for Dividing by a Polynomial Containing More Than One Term (Long Division):
| Step 1: | Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). |
| Step 2: | Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. |
| Step 3: | Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. |
| Step 4: | Subtract and bring down the next term. |
| Step 5: | Repeat Steps 2, 3, and 4 until there are no more terms to bring down. |
| Step 6: | Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. |
| Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). In this case, the problem is ready as is. |  |
| Step 2: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have x3 divided by x which is x2. |  |
| Step 3: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply x2 and x + 2. |  |
| Step 4: Subtract and bring down the next term. |  |
| Step 5: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have –6x2 divided by x which is –6x. |  |
| Step 6: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply –6x and x + 2. |  |
| Step 7: Subtract and bring down the next term. |  |
| Step 8: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have 14x divided by x which is +14. |  |
| Step 9: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply 14 and x + 2. |  |
| Step 10: Subtract and notice there are no more terms to bring down. |  |
| Step 11: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. |  |
Example 2 – Divide: 
| Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). In this case, we should get: |  |
| Step 2: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have 4x3 divided by x2 which is 4x. |  |
| Step 3: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply 4x and x2 + 3x – 2. |  |
| Step 4: Subtract and bring down the next term. |  |
| Step 5: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have –25x2 divided by x2 which is –25. |  |
| Step 6: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply –25 and x2+ 3x – 2. |  |
| Step 7: Subtract and notice there are no more terms to bring down. |  |
| Step 8: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. |  |
Example 3 – Divide: 
| Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). In this case, we should get: |  |
| Step 2: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have 3x3 divided by x2 which is 3x. |  |
| Step 3: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply 3x and x2 – 1. |  |
| Step 4: Subtract and bring down the next term. |  |
| Step 5: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. In this case, we have –2x2 divided by x2 which is –2. |  |
| Step 6: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we need to multiply –2 and x2 – 1. |  |
| Step 7: Subtract and notice there are no more terms to bring down. |  |
| Step 8: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. |  |
Example 4 – Divide: 
| Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). In this case, we should get: |  |
| Steps 2, 3, and 4: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Next multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we should get 2x3/2x = x2 and x2(2x + 3). Finally, subtract and bring down the next term. |  |
| Steps 5, 6, and 7: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Next multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we should get 4x2/2x = 2x and 2x(2x + 3). Finally, subtract and bring down the next term. |  |
| Steps 8, 9, and 10: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Next multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we should get –4x/2x = –2 and –2(2x + 3). Finally, subtract and notice there are no more terms to bring down. |  |
| Step 11: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. |  |
Example 5 – Divide: 
| Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). In this case, the problem is ready as is. |  |
| Steps 2, 3, and 4: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Next multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we should get x3/x2 = x and x(x2 + x – 6). Finally, subtract and bring down the next term. |  |
| Steps 5, 6, and 7: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Next multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. In this case, we should get 2x2/x2 = 2 and 2(x2 + x – 6). Finally, subtract and notice there are no more terms to bring down. |  |
| Step 8: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. In this case, there is no remainder, so you do not need to write the fraction. |  |
Source: http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/long_division/long_division.html
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