Factoring a Trinomial with a Leading Coefficient of 1
Learning Outcomes
- Factor a trinomial with a leading coefficient of [latex]1[/latex]
- Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[/latex]
Factoring Trinomials in the form [latex]x^{2}+bx+c[/latex]
To factor a trinomial in the form [latex]x^{2}+bx+c[/latex], find two integers, r and s, whose product is c and whose sum is b.[latex]\begin{array}{l}r\cdot{s}=c\\\text{ and }\\r+s=b\end{array}[/latex]
Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\left(x+r\right)[/latex] and [latex]\left(x+s\right)[/latex].Factors whose product is [latex]6[/latex] | Sum of the factors |
---|---|
[latex]1\cdot6=6[/latex] | [latex]1+6=7[/latex] |
[latex]2\cdot3=6[/latex] | [latex]2+3=5[/latex] |
Example
Factor [latex]x^{2}+5x+6[/latex].Answer: Use values from the chart above. Replace [latex]5x[/latex] with [latex]2x+3x[/latex].
[latex]x^{2}+2x+3x+6[/latex]
Group the pairs of terms.[latex]\left(x^{2}+2x\right)+\left(3x+6\right)[/latex]
Factor x out of the first pair of terms[latex]x\left(x+2\right)+\left(3x+6\right)[/latex]
Factor [latex]3[/latex] out of the second pair of terms.[latex]x\left(x+2\right)+3\left(x+2\right)[/latex]
Factor out [latex]\left(x+2\right)[/latex].[latex]\left(x+2\right)\left(x+3\right)[/latex]
Answer
[latex-display]\left(x+2\right)\left(x+3\right)[/latex-display]Factors whose product is [latex]−12[/latex] | Sum of the factors |
---|---|
[latex]1\cdot−12=−12[/latex] | [latex]1+−12=−11[/latex] |
[latex]2\cdot−6=−12[/latex] | [latex]2+−6=−4[/latex] |
[latex]3\cdot−4=−12[/latex] | [latex]3+−4=−1[/latex] |
[latex]4\cdot−3=−12[/latex] | [latex]4+−3=1[/latex] |
[latex]6\cdot−2=−12[/latex] | [latex]6+−2=4[/latex] |
[latex]12\cdot−1=−12[/latex] | [latex]12+−1=11[/latex] |
Example
Factor [latex]x^{2}+x–12[/latex].Answer: Rewrite the trinomial using the values from the chart above. Use values [latex]r=4[/latex] and [latex]s=−3[/latex].
[latex]x^{2}+4x+−3x–12[/latex]
Group pairs of terms.[latex]\left(x^{2}+4x\right)+\left(−3x–12\right)[/latex]
Factor [latex]x[/latex] out of the first group.[latex]x\left(x+4\right)+\left(-3x-12\right)[/latex]
Factor [latex]−3[/latex] out of the second group.[latex]x\left(x+4\right)–3\left(x+4\right)[/latex]
Factor out [latex]\left(x+4\right)[/latex].[latex]\left(x+4\right)\left(x-3\right)[/latex]
Answer
[latex-display]\left(x+4\right)\left(x-3\right)[/latex-display]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 [latex]x^{2}+bx+c[/latex], consider the following tips. Look at the c term first.- 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.
- 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.
- If the c term is positive and the b term is positive, then both r and s are positive.
- If the c term is positive and the b term is negative, then both r and s are negative.
- 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 [latex]|r|>|s|[/latex], then r is positive and s is negative.
- 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 [latex]|r|>|s|[/latex], then r is negative and s is positive.
Trinomial | [latex]x^{2}+7x+10[/latex] | [latex]x^{2}+5x+6[/latex] | [latex]x^{2}+x-12[/latex] |
---|---|---|---|
r and s values | [latex]r=+5,s=+2[/latex] | [latex]r=+2,s=+3[/latex] | [latex]r=+4,s=–3[/latex] |
Factored form | [latex]\left(x+5\right)\left(x+2\right)[/latex] | [latex]\left(x+2\right)\left(x+3\right)[/latex] | [latex](x+4)(x–3)[/latex] |
The Shortcut
Notice that in each of the examples above, 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 [latex]x^{2}+bx+c[/latex] (where the coefficient in front of [latex]x^{2}[/latex] is [latex]1[/latex]), if you can identify the correct r and s values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let's look at some examples where we use this idea.[latex]\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)[/latex]
Example
Factor: [latex]y^2+6y-27[/latex]Answer: Find r and s:
Factors whose product is -27 | Sum of the factors |
---|---|
[latex]1\cdot{-27}=-27[/latex] | [latex]1-27=-26[/latex] |
[latex]3\cdot{-9}=-27[/latex] | [latex]3-9=-6[/latex] |
[latex]-3\cdot{9}=-27[/latex] | [latex]-3+9=6[/latex] |
[latex]\begin{array}{l}r=-3\\s=9\end{array}[/latex]
It helps to start by writing two empty sets of parentheses:[latex]\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)[/latex]
The squared term is y, so we will place a y in each set of parentheses:
[latex]\left(y\,\,\,\,\,\,\,\,\right)\left(y\,\,\,\,\,\,\,\,\right)[/latex]
Now we can fill in the rest of each binomial with the values we found for r and s.
[latex]\left(y-3\right)\left(y+9\right)[/latex]
Note how we kept the sign on each of the values. The nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.
[latex]\begin{array}{l}\left(y-3\right)\left(y+9\right)\\=y^2+9y-3y-27\\=y^2+6y-27\end{array}[/latex]
Answer
[latex-display]\left(y-3\right)\left(y+9\right)[/latex-display]Example
Factor: [latex]-m^2+16m-48[/latex]Answer: There is a negative in front of the squared term, so we will factor out a negative one from the whole trinomial first. Remember, this boils down to changing the sign of all the terms:
[latex]-m^2+16m-48=-1\left(m^2-16m+48\right)[/latex]
Now we can factor [latex]\left(m^2-16m+48\right)[/latex] by finding r and s. Note that b is negative, and c is positive so we are probably looking for two negative numbers:
Factors whose product is 48 | Sum of the factors |
---|---|
[latex]-1\cdot{-48}=48[/latex] | [latex]-1-48=-49[/latex] |
[latex]-2\cdot{-12}=48[/latex] | [latex]-2-12=-14[/latex] |
[latex]-3\cdot{-16}=-48[/latex] | [latex]-3-16=-19[/latex] |
[latex]-4\cdot{-12}=-48[/latex] | [latex]-4-12=-16[/latex] |
There are more factors whose product is [latex]48[/latex], but we have found the ones that sum to [latex]-16[/latex], so we can stop.
[latex]\begin{array}{l}r=-4\\s=-12\end{array}[/latex]
Now we can fill in each binomial with the values we found for r and s, make sure to use the correct variable!
[latex]\left(m-4\right)\left(m-12\right)[/latex]
We are not done yet, remember that we factored out a negative sign in the first step. We need to remember to include that.
[latex]-1\left(m-4\right)\left(m-12\right)[/latex]
Answer
[latex-display]-m^2+16m-48=-1\left(m-4\right)\left(m-12\right)[/latex-display]Try It
[ohm_question]7900[/ohm_question]How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[/latex], factor it
- List factors of [latex]c[/latex].
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]c[/latex] with a sum of [latex]b[/latex].
- Write the factored expression [latex]\left(x+p\right)\left(x+q\right)[/latex].
Example
Factor [latex]x^{2}+x–12[/latex].Answer: Consider all the combinations of numbers whose product is [latex]-12[/latex] and list their sum.
Factors whose product is [latex]−12[/latex] | Sum of the factors |
---|---|
[latex]1\cdot−12=−12[/latex] | [latex]1+−12=−11[/latex] |
[latex]2\cdot−6=−12[/latex] | [latex]2+−6=−4[/latex] |
[latex]3\cdot−4=−12[/latex] | [latex]3+−4=−1[/latex] |
[latex]4\cdot−3=−12[/latex] | [latex]4+−3=1[/latex] |
[latex]6\cdot−2=−12[/latex] | [latex]6+−2=4[/latex] |
[latex]12\cdot−1=−12[/latex] | [latex]12+−1=11[/latex] |
[latex]\left(x+4\right)\left(x-3\right)[/latex]
Think About It
Which property of multiplication can be used to describe why [latex]\left(x+4\right)\left(x-3\right) =\left(x-3\right)\left(x+4\right)[/latex]. Use the textbox below to write down your ideas before you look at the answer. [practice-area rows="2"][/practice-area]Answer: The commutative property of multiplication states that factors may be multiplied in any order without affecting the product.
Example
Factor [latex]{x}^{2}-7x+6[/latex].Answer: List the factors of [latex]6[/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.
Factors of [latex]6[/latex] | Sum of Factors |
---|---|
[latex]1,6[/latex] | [latex]7[/latex] |
[latex]2, 3[/latex] | [latex]5[/latex] |
[latex]-1, -6[/latex] | [latex]-7[/latex] |
[latex]-2, -3[/latex] | [latex]-5[/latex] |
Think About It
Can every trinomial be factored as a product of binomials? Mathematicians often use a counterexample to prove or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient [latex]1[/latex] that cannot be factored as a product of binomials? Use the textbox below to write your ideas. [practice-area rows="2"][/practice-area]Answer: Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. A counterexample would be: [latex]x^2+3x+7[/latex]
Try It
[ohm_question]7907[/ohm_question]Contribute!
Licenses & Attributions
CC licensed content, Original
- Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.