7.2: Factoring trinomials of the form x² + bx + c
- Page ID
- 45067
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Factoring with three terms, or trinomials, is the most important technique, especially in further algebra. Since factoring is a product of factors, we first look at multiplying to develop the process of factoring trinomials.
Factoring Trinomials of the Form \(x^2+bx+c\)
If we multiply \((x + p)(x + q)\), we would get
\[\begin{array}{c}x^2+px+qx+pq \\ x^2+(p+q)x+pq\end{array}\nonumber\]
Notice the two factors of the last coefficient must add up to be the middle coefficient, i.e.,
\[p\cdot q=c\text{ and }p+q=b\nonumber\]
Hence, if we can find two numbers whose sum is \(b\) and that multiply to \(c\), then we can split the middle term and factor by grouping.
Step 1. Find two numbers, \(p\) and \(q\), whose sum is \(b\) and product is \(c\).
Step 2. Rewrite the expression so that the middle term is split into two terms, \(p\) and \(q\).
Step 3. Factor by grouping.
Step 4. Verify the factored form by finding the product.
Factor: \(x^2+9x+18\)
Solution
First we identify \(b = 9\) and \(c = 18\). We ask ourselves,“ What two numbers multiply to \(18\) that add up to \(9\)?”
Step 1. Find two numbers whose sum is \(9\) and product is \(18\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(2,9\) | \(18\) | \(11\) |
| \(3,6\) | \(18\) | \(9\) |
| \(1,18\) | \(18\) | \(19\) |
We can see from the table that \(3\) and \(6\) are the two numbers whose product is \(18\) and sum is \(9\). We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, \(3x\) and \(6x\): \[\begin{array}{c}x^2+9x+18 \\ x^2\color{blue}{\underset{\text{sum is }9x}{\underbrace{+3x+6x}}}\color{black}{}+18\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}x^2+6x+3x+18&\text{Group the first two terms and the last two terms} \\ (x^2+3x)+(6x+18)&\text{Factor }x\text{ from the first group and }6 \\ &\text{from the second group} \\ x(x+3)+6(x+3)&\text{Factor the GCF }(x+3) \\ (x+3)(x+6)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(x+3)(x+6)&\text{FOIL} \\ x^2+6x+3x+18&\text{Combine like terms} \\ x^2+9x+18&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \((x+3)(x+6)\).
Factor: \(x^2 − 4x + 3\)
Solution
First we identify \(b = −4\) and \(c = 3\). We ask ourselves,“ What two numbers multiply to \(3\) that add up to \(−4\)?”
Step 1. Find two numbers whose sum is \(−4\) and product is \(3\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(1,3\) | \(3\) | \(4\) |
| \(-1,-3\) | \(3\) | \(-4\) |
We can see from the table that \(−1\) and \(−3\) are the two numbers whose product is \(3\) and sum is \(−4\). We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, \(−1x\) and \(−3x\): \[\begin{array}{c}x^2-4x+3 \\ x^2\color{blue}{\underset{\text{sum is }-4x}{\underbrace{-1x-3x}}}\color{black}{}+3\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}x^2-1x-3x+3&\text{Group the first two terms and the last two terms} \\ (x^2-1x)+(-3x+3)&\text{Factor }x\text{ from the first group and }-3 \\ &\text{from the second group} \\ x(x-1)-3(x-1)&\text{Factor the GCF }(x-1) \\ (x-1)(x-3)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(x-1)(x-3)&\text{FOIL} \\ x^2-1x-3x+3&\text{Combine like terms} \\ x^2-4x+3&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \((x-1)(x+3)\).
Factor: \(x^2 − 8x − 20\)
Solution
First we identify \(b = −8\) and \(c = −20\). We ask ourselves,“ What two numbers multiply to \(−20\) that add up to \(−8\)?”
Step 1. Find two numbers whose sum is \(−8\) and product is \(−20\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(-4,5\) | \(-20\) | \(1\) |
| \(4,-5\) | \(-10\) | \(-1\) |
| \(-2,10\) | \(-20\) | \(8\) |
| \(2,-10\) | \(-20\) | \(-8\) |
| \(1,-20\) | \(-20\) | \(-19\) |
| \(-1,20\) | \(-20\) | \(19\) |
We can see from the table that \(2\) and \(−10\) are the two numbers whose product is \(−20\) and sum is \(−8\). We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, \(2x\) and \(−10x\): \[\begin{array}{c}x^2-8x-20 \\ x^2\color{blue}{\underset{\text{sum is }-8x}{\underbrace{+2x-10x}}}\color{black}{}-20\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}x^2+2x-10x-20&\text{Group the first two terms and the last two terms} \\ (x^2+2x)+(-10x-20)&\text{Factor }x\text{ from the first group and }-10 \\ &\text{from the second group} \\ x(x+2)-10(x+2)&\text{Factor the GCF }(x+2) \\ (x+2)(x-10)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(x+2)(x-10)&\text{FOIL} \\ x^2+2x-10x-20&\text{Combine like terms} \\ x^2-8x-20&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \((x+2)(x+10)\).
Factor: \(a^2-9ab+14b^2\)
Solution
First we identify \(b = −9\) and \(c = 14\). We ask ourselves,“ What two numbers multiply to \(14\) that add up to \(−9\)?”
Step 1. Find two numbers whose sum is \(−9\) and product is \(14\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(2,7\) | \(14\) | \(9\) |
| \(-2,-7\) | \(14\) | \(-9\) |
| \(-1,-14\) | \(14\) | \(-15\) |
| \(1,14\) | \(14\) | \(15\) |
We can see from the table that \(−2\) and \(−7\) are the two numbers whose product is \(14\) and sum is \(−9\). We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, \(−2ab\) and \(−7ab\): \[\begin{array}{c}a^2-9ab+14b^2 \\ a^2\color{blue}{\underset{\text{sum is }-9ab}{\underbrace{-2ab-7ab}}}\color{black}{}+14b^2\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}a^2-2ab-7ab+14b^2&\text{Group the first two terms and the last two terms} \\ (a^2-2ab)+(-7ab+14b^2)&\text{Factor }a\text{ from the first group and }-7b \\ &\text{from the second group} \\ a(a-2b)-7b(a-2b)&\text{Factor the GCF }(a-2b) \\ (a-2b)(a-7b)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(a-2b)(a-7b)&\text{FOIL} \\ a^2-2ab-7ab+14b^2&\text{Combine like terms} \\ a^2-9ab+14b^2&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \((a-2b)(a-7b)\).
There is a shortcut for factoring expressions of the type \(x^2 +bx+c\). Once we identify the two numbers, \(p\) and \(q\), whose product is \(c\) and sum is \(b\), we can see these two numbers are the numbers in the factored form, i.e., \((x + p)(x + q)\). We can use this shortcut only when the coefficient of \(x^2\) is \(1\). (We discuss when the coefficient is a number other than \(1\) in the next section.)
Factor: \(x^2-7x-18\)
Solution
First we identify \(b = −7\) and \(c = −18\). We ask ourselves,“ What two numbers multiply to \(−18\) that add up to \(−7\)?”
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(-2,9\) | \(-18\) | \(7\) |
| \(2,-9\) | \(-18\) | \(-7\) |
| \(-1,18\) | \(-18\) | \(17\) |
| \(1,-18\) | \(-18\) | \(-17\) |
We can see from the table that \(2\) and \(−9\) are the two numbers whose product is \(−18\) and sum is \(−7\). We use these two numbers to rewrite the expression in factored form:
\[\begin{array}{c}x^2-7x-18 \\ (x\color{blue}{+2}\color{black}{})(x\color{blue}{-9}\color{black}{})\end{array}\nonumber\]
We can always verify the factored form by multiplying and obtaining the original expression.
Factor: \(m^2-mn-30n^2\)
Solution
First we identify \(b = −1\) and \(c = −30\). We ask ourselves,“ What two numbers multiply to \(−30\) that add up to \(−1\)?”
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(-2,15\) | \(-30\) | \(13\) |
| \(2,-15\) | \(-30\) | \(-13\) |
| \(5,-6\) | \(-30\) | \(-1\) |
| \(-5,6\) | \(-30\) | \(1\) |
| \(1,-30\) | \(-30\) | \(-29\) |
| \(-1,30\) | \(-30\) | \(29\) |
We can see from the table that \(5\) and \(−6\) are the two numbers whose product is \(−30\) and sum is \(−1\). We use these two numbers to rewrite the expression in factored form:
\[\begin{array}{c}m^2-mn-30n^2 \\ (m\color{blue}{+5n}\color{black}{})(m\color{blue}{-6n}\color{black}{})\end{array}\nonumber\]
We can always verify the factored form by multiplying and obtaining the original expression.
Factor: \(x^2+2x+6\)
Solution
First we identify \(b = 2\) and \(c = 6\). We ask ourselves,“ What two numbers multiply to \(6\) that add up to \(2\)?”
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(2,3\) | \(6\) | \(5\) |
| \(-2,-3\) | \(6\) | \(-5\) |
| \(1,6\) | \(6\) | \(7\) |
| \(-1,-6\) | \(6\) | \(-7\) |
We can see from the table that there aren’t any factors of \(6\) whose sum is \(2\). We only obtain sums with \(5\) and \(7\)’s. In this case, we call this trinomial not factorable, or better yet, the trinomial is prime.
If a trinomial (or polynomial) is not factorable, then we say we the trinomial is prime.
Factoring Trinomials of the Form \(x^2+bx+c\) with a Greatest Common Factor
Factoring the GCF is always the first step in factoring expressions. If all terms have a common factor, we, first, factor the GCF and then factor as usual.
Factor: \(3x^2-24x+45\)
Solution
Notice all three terms have a common factor of \(3\). We factor a \(3\) first, then factor as usual.
\[\begin{array}{rl}3x^2-24x+45&\text{Factor the GCF} \\ \color{blue}{3}\color{black}{}(x^2-8x+15)\end{array}\nonumber\]
Next, we only concentrate on the expression in the parenthesis. What two numbers multiply to \(15\) that add up to \(−8\)?
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(3,5\) | \(15\) | \(8\) |
| \(-3,-5\) | \(15\) | \(-8\) |
We can see from the table that \(−3\) and \(−5\) are the two numbers whose product is \(15\) and sum is \(−8\). We use these two numbers to rewrite the expression in factored form:
\[\begin{array}{c}3x^2-24x+4 \\ 3(x\color{blue}{-3}\color{black}{})(x\color{blue}{-5}\color{black}{})\end{array}\nonumber\]
We can always verify the factored form by multiplying and obtaining the original expression.
Students tend to forget to write the GCF in the final answer. Be sure to always include the GCF in the final factored form.
Also, to factor completely, it is required the GCF is factored out of the expression. If not, then the expression is not factored completely.
The first person to use letters for unknown values was Francois Vieta in 1591 in France. He used vowels to represent variables for solving, just as codes used letters to represent an unknown message.
Factoring Trinomials of the Form \(x^2+bx+c\) Homework
Factor completely.
\(p^2+17p+72\)
\(n^2-9n+8\)
\(x^2-9x-10\)
\(b^2+12b+32\)
\(x^2+3x-70\)
\(n^2-8n+15\)
\(p^2+15p+54\)
\(n^2-15n+56\)
\(u^2-8uv+15v^2\)
\(m^2+2mn-8n^2\)
\(x^2-11xy+18y^2\)
\(x^2+xy-12y^2\)
\(x^2+4xy-12y^2\)
\(5a^2+60a+100\)
\(6a^2+24a-192\)
\(6x^2+18xy+12y^2\)
\(6x^2+96xy+378y^2\)
\(x^2+x-72\)
\(x^2+x-30\)
\(x^2+13x+40\)
\(b^2-17b+70\)
\(x^2+3x-18\)
\(a^2-6a-27\)
\(p^2+7p-30\)
\(m^2-15mn+50n^2\)
\(m^2-3mn-40n^2\)
\(x^2+10xy+16y^2\)
\(u^2-9uv+14v^2\)
\(x^2+14xy+45y^2\)
\(4x^2+52x+168\)
\(5n^2-45n+40\)
\(5v^2+20v-25\)
\(5m^2+30mn-90n^2\)
\(6m^2-36mn-162n^2\)