7.3: Factoring trinomials of the form ax² + bx + c
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When factoring trinomials, we factored by grouping after we split the middle term. We continue to use this method for further factoring, like trinomials of the form \(ax^2 + bx + c\), where \(a,\: b,\) and \(c\) are coefficients.
French philosopher Rene Descartes first used letters from the beginning of the alphabet to represent known values, \(a,\: b,\: c,\) and letters from the end to represent unknown values, \(x,\: y,\: z\).
A simple example of a trinomial of the form \(ax^2 + bx + c\), where \(a\neq 1\), is \(3x^2 + 11x + 6\). Recall, that this trinomial was a product of two binomials and we applied the FOIL method to obtain the result. Let’s take a look at the binomial factors:
\[(3x+2)(x+3)=\color{blue}{\underset{\text{F}}{\underbrace{3x^2}}}\color{black}{}+\color{blue}{\underset{\text{O}}{\underbrace{9x}}}\color{black}{}+\color{blue}{\underset{\text{I}}{\underbrace{2x}}}\color{black}{}+\color{blue}{\underset{\text{L}}{\underbrace{6}}}\color{black}{}=3x^2+11x+6\nonumber\]
In this section, we reverse the FOIL method and factor by grouping, or use the method of trial-and-error.
Factoring Trinomials of the Form ax² + bx + c Using Grouping
ax² + bx + c
Step 1. Find two numbers, \(p\) and \(q\), whose sum is \(b\) and product is \(a\cdot c\).
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: \(3x^2+11x+6\)
Solution
First we identify \(a = 3,\: b = 11\) and \(c = 6\). We ask ourselves,“ What two numbers multiply to \(3\cdot 6\) that add up to \(11\)?”
Step 1. Find two numbers whose sum is \(11\) 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 \(2\) and \(9\) are the two numbers whose product is \(18\) and sum is \(11\). 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 \(9x\): \[\begin{array}{c}3x^2+11x+6 \\ 3x^2\color{blue}{\underset{\text{sum is }11x}{\underbrace{+2x+9x}}}\color{black}{}+6\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}3x^2+2x+9x+6&\text{Group the first two terms and the last two terms} \\ (3x^2+2x)+(9x+6)&\text{Factor }x\text{ from the first group and }3 \\ &\text{from the second group} \\ x(3x+2)+3(3x+2)&\text{Factor the GCF }(3x+2) \\ (3x+2)(x+3)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(3x+2)(x+3)&\text{FOIL} \\ 3x^2+2x+9x+6&\text{Combine like terms} \\ 3x^2+11x+6&\checkmark\text{Original expression}\end{array}\nonumber\]
Thus, the factored form is \((3x+2)(x+3)\).
Factor: \(8x^2-2x-15\)
Solution
First we identify \(a = 8\), \(b = −2\) and \(c = −15\). We ask ourselves,“ What two numbers multiply to \(8\cdot −15\) that add up to \(−2\)?”
Step 1. Find two numbers whose sum is \(−2\) and product is \(−120\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(-8,15\) | \(-120\) | \(7\) |
| \(8,-15\) | \(-120\) | \(-7\) |
| \(-10,12\) | \(-120\) | \(2\) |
| \(10,-12\) | \(-120\) | \(-2\) |
We can see from the table that \(10\) and \(−12\) are the two numbers whose product is \(−120\) and sum is \(−2\). We use these two numbers in Step 2.
Step 2. Rewrite the expression so that the middle term is split into two terms, \(10x\) and \(−12x\): \[\begin{array}{c}8x^2-2x-15 \\ 8x^2\color{blue}{\underset{\text{sum is }-2x}{\underbrace{+10x-12x}}}\color{black}{}-15\end{array}\nonumber\]
Step 3. Factor by grouping. \[\begin{array}{rl}8x^2+10x-12x-15&\text{Group the first two terms and the last two terms} \\ (8x^2+10x)+(-12x-15)&\text{Factor }2x\text{ from the first group and }-3 \\ &\text{from the second group} \\ 2x(4x+5)-3(4x+5)&\text{Factor the GCF }(4x+5) \\ (4x+5)(2x-3)&\text{Factored form}\end{array}\nonumber\]
Step 4. Verify the factored form by finding the product: \[\begin{array}{rl}(4x+5)(2x-3)&\text{FOIL} \\ 8x^2+10x-12x-15&\text{Combine like terms} \\ 8x^2-2x-15&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \((4x+5)(2x-3)\).
When factoring trinomials of the form \(ax^2 + bx + c\), there is no shortcut as we saw when we factored trinomials of the form \(x^2 + bx + c\). There is no way around this despite student efforts and desires for a shortcut. The closest we get to a shortcut is factoring by trial-and-error in which next we discuss.
Since \(a\neq 1\), this makes factoring more involved and forces us to factor by grouping, sometimes called the ac method, or by trial-and-error.
Factoring Trinomials of the Form \(ax^2+bx+c\) Using Trial-and-Error
Factor: \(10x^2-27x+5\)
Solution
Factoring by trial-and-error is just that- trial and error. Recall, FOIL: first, outer, inner, last. We know the first’s product is \(10x^2\) and the last’s product is \(5\). Let’s try different combinations and FOIL each combination to obtain the original expression. This sometimes works out faster than the \(ac\) method and sometimes not. Hence, trial and error.
| binomials | FOIL | Yes or no? |
|---|---|---|
| \((5x+5)(2x+1)\) | \(10x^2+15x+5\) | NO |
| \((2x+5)(5x+1)\) | \(10x^2+27x+5\) | NO |
| \((2x+5)(5x-1)\) | \(10x^2+23x-5\) | NO |
| \((2x-5)(5x-1)\) | \(10x^2-27x+5\) | YES |
Once we obtain the original expression we can stop taking combinations. We have found the factored form of the original expression:
\[(2x-5)(5x-1)\nonumber\]
We can see that this method may not be the most time efficient unless, of course, we get lucky and obtain the the right combination quickly.
For factoring trinomials of the form \(x^2 + bx + c\), here are some helpful tips when factoring by trialand-error:
- Always look at the \(x^2\) and \(c\) terms first and determine factors of these terms.
- According to the trinomial’s signs of the terms, the binomial factors take on these signs:
| trinomials | binomials | signs in binomial factors |
|---|---|---|
| \(x^2+mx+n\) | \((x+p)(x+q)\) | all positive |
| \(x^2-mx+n\) | \((x-p)(x-q)\) | all negative |
| \(x^2-mx-n\) | \((x-p)(x+q)\) | one positive, one negative |
| \(x^2+mx-n\) | \((x-p)(x+q)\) | one positive, one negative |
Now, for the last two cases, the binomials with subtraction and addition is to be determined by the student when FOIL-ing each combination. Also note, \(m,\: n,\: p,\: q\) are all coefficients and numbers, respectively. In general, we can use these patterns of the signs in binomial factors for any trinomial form expression.
Factor: \(4x^2-xy-5y^2\)
Solution
We know the first’s product is \(4x^2\) and the last’s product is \(−5y^2\). Let’s try different combinations and FOIL each combination to obtain the original expression. Since the signs of the last two terms are negative, this means we have case \(3\) from above and the binomial factors will have alternating signs.
| binomials | FOIL | Yes or no? |
|---|---|---|
| \((2x+5y)(2x-y)\) | \(4x^2+8xy-5y^2\) | NO |
| \((2x-5y)(2x+y)\) | \(4x^2-8xy-5y^2\) | NO |
| \((4x+5y)(x-y)\) | \(4x^2+xy-5y^2\) | NO |
| \((4x-5y)(x+y)\) | \(4x^2-xy-5y^2\) | YES |
Once we obtain the original expression we can stop taking combinations. We have found the factored form of the original expression:
\[(4x-5y)(x+y)\nonumber\]
Factoring Trinomials of the Form \(ax^2+bx+c\) with a Greatest Common Factor
As always, when factoring, we will first look for a GCF, then factor as usual.
Factor: \(-18x^3-33x^2+30x\)
Solution
Notice all three terms have a common factor of \(−3x\). We factor \(−3x\) first, then factor as usual.
\[\begin{array}{rl}-18x^3-33x^2+30x&\text{Factor the GCF} \\ \color{blue}{-3x}\color{black}{}(6x^2+11x-10)\end{array}\nonumber\]
Next, we only concentrate on the expression in the parenthesis. Let’s factor by trial-and-error. We know the first’s product is \(6x^2\) and the last’s product is \(−10\). Let’s try different combinations and FOIL each combination to obtain the original expression. Since the signs of the last two terms are positive and negative, respectively, this means we have case \(4\) from above and the binomial factors will have alternating signs.
| binomials | FOIL | Yes or no? |
|---|---|---|
| \((2x+5)(3x-2)\) | \(4x^2+11x-10\) | NO |
| \((2x-5)(3x+2)\) | \(4x^2-11x-10\) | YES |
Once we obtain the original expression we can stop taking combinations. We have found the factored form of the original expression:
\[-3x(2x-5)(3x+2)\nonumber\]
Recall, we include the GCF in the final answer or else the trinomial isn’t factored completely. Note, if the \(ax^2\) term is negative, then we always factor a negative because factoring becomes less challenging when the leading term is positive.
Note
In general, when given a trinomial where the leading coefficient is negative, like in Example 7.3.5 , we treat the leading coefficient as a GCF and factor it out prior to factoring the trinomial, i.e., if \(a < 0\), then factor the negative out prior to factoring by grouping or trial-and-error.
Example 7.3.6
Factor: \(3x^2+2x-7\)
Solution
Let’s try to factor by the \(ac\) method. First we identify \(a = 3,\: b = 2\) and \(c = −7\). We ask ourselves,“ What two numbers multiply to \(3\cdot −7\) that add up to \(2\)?”
Step 1. Find two numbers whose sum is \(2\) and product is \(−21\):
| \(p\) and \(q\) | Product | Sum |
|---|---|---|
| \(-3,7\) | \(-21\) | \(4\) |
| \(3,-7\) | \(-21\) | \(-4\) |
| \(-1,21\) | \(-21\) | \(20\) |
| \(1,-21\) | \(-21\) | \(-20\) |
We can see from the table that there aren’t any factors of \(−21\) whose sum is \(2\). We only obtain sums with \(4\) and \(20\)’s. In this case, we call this trinomial prime.
Factoring Trinomials Using Substitution
Even though we are used to factoring expressions of the form \(ax^2 +bx+c\), we can apply these same factoring strategies to other trinomials of the same form where
\[a(\text{base})^2+b(\text{base})+c\nonumber\]
Example 7.3.7
For example, factor \(2(y+1)^2+3(y+1)-35\).
Solution
We can see that the base \((y + 1)\) is squared, then to the first power in the linear term. Let’s allow \(\color{blue}{u}\color{black}{} = y + 1\) and rewrite the expression in terms of \(u\): \(\underset{\color{blue}{u}}{\underbrace{2(y+1)^2}}\color{black}{}+\underset{\color{blue}{u}}{\underbrace{3(y+1)}}\color{black}{}-35\)
\[2\color{blue}{u}\color{black}{}^2+3\color{blue}{u}\color{black}{}-35\nonumber\]
Let’s factor the expression in terms of \(u\). Then substitute back \(u = y + 1\) in the end.
\[\begin{array}{rl}2u^2+3u-35&\text{Factor using trial-and-error} \\ (2u-\quad )(u+\quad )&\text{Using the last case in trial-and-error,} \\ &\text{we know the signs will be }-,+ \\ (2u-7)(u+5)&\text{Two factors whose product is }-35 \\ &\text{are }-7,5\end{array}\nonumber\]
Checking with FOIL:
\[\begin{array}{c}2u^2+10u-7u-35 \\ 2u^2+3u-35\:\checkmark\end{array}\nonumber\]
Recall, \((2u − 7)(u + 5)\) is the factored form of the expression \(2u^2 + 3u − 35\), not the original expression, \(2(y + 1)^2 + 3(y + 1) − 35\). We used u-substitution to write the original expression in terms of \(u\) so we could easily factor. Lastly, we need to substitute back in \(\color{blue}{u}\color{black}{} = y + 1\) into the factored expression \((2\color{blue}{u}\color{black}{} − 7)(\color{blue}{u}\color{black}{} + 5)\) and simplify:
\[\begin{array}{c}\left(2\color{blue}{(y+1)}\color{black}{}-7\right) (\color{blue}{(y+1)}\color{black}{}+5) \\ (2y+2-7)(y+1+5)\end{array}\nonumber\]
Simplifying each factor, we get the final factored expression in terms of the original variable \(y\) to be
\[(2y-5)(y+6)\nonumber\]
Factoring expressions of the form \(a(\text{base})^2+b(\text{base})+c\)
When given a polynomial expression of the form \(a(\text{base})^2 + b(\text{base}) + c\), we can use the same factoring strategies to factor these expressions by using u-substitution, where \[u=\text{base}\nonumber\]
Note
When using u-substitution, be sure to substitute back in the original base into the factored expression so that the final factored expression contains the original variable.
Example 7.3.8
Factor: \(z^{2/3}+2z^{1/3}-80\)
Solution
We can see that the base is \(z\). Furthermore, we can rewrite the first term as \(\left(z^{1/3}\right)^{2}\), and the second term as \(\left(z^{1/3}\right)^{1}=z^{1/3}\). Let's allow \(\color{blue}{u}\color{black}{}=z^{1/3}\) and rewrite the expression in terms of \(u\):
\[\begin{array}{c}\underset{\color{blue}{u}}{\underbrace{\left(z^{1/3}\right)^2}}\color{black}{}+\underset{\color{blue}{u}}{\underbrace{2z^{1/3}}}\color{black}{}-80 \\ \color{blue}{u}\color{black}{}^2+2\color{blue}{u}\color{black}{}-80\end{array}\nonumber\]
Let’s factor the expression in terms of \(u\). Then substitute back \(u = z^{1/3}\) in the end.
\[\begin{array}{rl}u^2+2u-80&\text{Factor using trial-and-error} \\ (u-\quad )(u+\quad )&\text{Using the last case in trial-and-error} \\ &\text{we know the signs will be }-,+ \\ (u-8)(u+10)&\text{Two factors whose product is }-80 \\ &\text{are }-8,10\end{array}\nonumber\]
Checking with FOIL:
\[\begin{array}{c}u^2+10u-8u-80 \\ u^2+2u-80\:\checkmark\end{array}\nonumber\]
Recall, \((u − 8)(u + 10)\) is the factored form of the expression \(u^2 + 2u − 80\), not the original expression, \(z^{2/3} + 2z^{1/3} − 80\). Lastly, we need to substitute back in \(\color{blue}{u}\color{black}{} = z^{1/3}\) into the factored expression \((\color{blue}{u}\color{black}{} − 8)(\color{blue}{u}\color{black}{} + 10)\) and simplify:
\[\left(\color{blue}{z^{1/3}}\color{black}{}-8\right)\left(\color{blue}{z^{1/3}}\color{black}{}+10\right)\nonumber\]
Simplifying each factor, we get the final factored expression in terms of the original variable \(z\) to be
\[\left(z^{1/3}-8\right) \left(z^{1/3}+10\right)\nonumber\]
Factoring Trinomials of the Form \(ax^2+bx+c\) Homework
What are possible values of \(p\) given the trinomial \(ax^2+bx+c\)?
Exercise 7.3.1
\((2x+p)(x+3)\) given \(2x^2+7x+3\)
Exercise 7.3.2
\((3x+p)(5x+2)\) given \(15x^2-14x-8\)
Exercise 7.3.3
\((x+p)(4x-3)\) given \(4x^2+x-3\)
Exercise 7.3.4
\((7x+p)(x-4)\) given \(7x^2-30x+8\)
Factor completely by grouping.
Exercise 7.3.5
\(7x^2-48x+36\)
Exercise 7.3.6
\(7b^2+15b+2\)
Exercise 7.3.7
\(5a^2-13a-28\)
Exercise 7.3.8
\(2x^2-5x+2\)
Exercise 7.3.9
\(2x^2+19x+35\)
Exercise 7.3.10
\(2b^2-b-3\)
Exercise 7.3.11
\(5k^2+13k+6\)
Exercise 7.3.12
\(3x^2-17x+20\)
Exercise 7.3.13
\(3x^2+17xy+10y^2\)
Exercise 7.3.14
\(5x^2+28xy-49y^2\)
Exercise 7.3.15
\(6x^2-39x-21\)
Exercise 7.3.16
\(21k^2-87k-90\)
Exercise 7.3.17
\(14x^2-60x+16\)
Exercise 7.3.18
\(6x^2+29x+20\)
Exercise 7.3.19
\(4k^2-17k+4\)
Exercise 7.3.20
\(4x^2+9xy+2y^2\)
Factor completely by trial-and-error.
Exercise 7.3.21
\(4m^2-9mn-9n^2\)
Exercise 7.3.22
\(4x^2+13xy+3y^2\)
Exercise 7.3.23
\(12x^2+62xy+70y^2\)
Exercise 7.3.24
\(24x^2-52xy+8y^2\)
Exercise 7.3.25
\(7n^2-44n+12\)
Exercise 7.3.26
\(7v^2-24v-16\)
Exercise 7.3.27
\(5n^2-4n-20\)
Exercise 7.3.28
\(3r^2-4r-4\)
Exercise 7.3.29
\(7x^2+29x-30\)
Exercise 7.3.30
\(5k^2-26k+24\)
Exercise 7.3.31
\(3r^2+16r+21\)
Exercise 7.3.32
\(3u^2+13uv-10v^2\)
Exercise 7.3.33
\(7x^2-2xy-5y^2\)
Exercise 7.3.34
\(5u^2+31uv-28v^2\)
Exercise 7.3.35
\(10a^2-54a-36\)
Exercise 7.3.36
\(21n^2+45n-54\)
Factor completely.
Exercise 7.3.37
\(4r^2+r-3\)
Exercise 7.3.38
\(6p^2+11p-7\)
Exercise 7.3.39
\(4r^2+3r-7\)
Exercise 7.3.40
\(4m^2+6mn+6n^2\)
Exercise 7.3.41
\(4x^2-6xy+30y^2\)
Exercise 7.3.42
\(18u^2-3uv-36v^2\)
Exercise 7.3.43
\(16x^2+60xy+36y^2\)
Exercise 7.3.44
\(12x^2+50xy+28y^2\)
Factor completely by using substitution.
Exercise 7.3.45
\((x-2y)^2+7(x-2y)-18\)
Exercise 7.3.46
\(3(a-b)^2-(a-b)-44\)
Exercise 7.3.47
\((5a-3b)^2+8(5a-3b)+16\)
Exercise 7.3.48
\((x-4y)^2-10(x-4y)+25\)
Exercise 7.3.49
\((3r-4)^2-4(3r-4)-12\)
Exercise 7.3.50
\((7m-1)^2+12(7m-1)-45\)
Exercise 7.3.51
\(w^{2/5}-2w^{1/5}-80\)
Exercise 7.3.52
\(x-5x^{1/2}-50\)
Exercise 7.3.53
\(x^{2/3}+8x^{1/3}-20\)
Exercise 7.3.54
\(x^{6/7}-10x^{3/7}+21\)