# 2.9: Factoring ax² + bx + c when a =1

- Page ID
- 92368

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we concentrate on learning how to factor trinomials having the form \(ax^2 +bx+c\) when \(a = 1\). Let's begin by looking at some patterns of multiplication of binomials:

\[\begin{align*}(x+2)(x+3)&=x^2+5x+6\\[4pt] (x-2)(x-3)&=x^2-5x+6\\[4pt] (x+2)(x-3)&=x^2-x-6\\[4pt] (x-2)(x+3)&=x^2+x-6 \end{align*}\]

Notice that when we multiply two binomials together that have the same types of terms (namely x-terms and constants in the examples above) that the result is a trinomial. This will always be the case when we multiply two binomials that have the same types of terms.

\[\begin{align*} (y^2+4)(y^2+2)&=y^4+6y^2+8 \\[4pt] (b^3+5)(b^3-9)&=b^6-4b^3-45\\[4pt] (kl+5)(kl-10)&=k^2l^2-5kl-50 \\[4pt] (m^4n-3)(m^4n-7)&=m^8n^2-10m^4n+21\end{align*}\]

Also, notice that the trinomial we get when we multiply has two terms that are the same as each term of the binomials and that the third term of the trinomial has variables that are a product of the variable terms of the binomial.

\[\begin{align*} (x+5)(x-3)&=x^2+2x-15 \quad{\color{Red}(\text{notice } x \cdot x=x^2)}\\[4pt] (y^2+4)(y^2+2)&=y^4+6y^2+8 \quad{\color{Red}(\text{notice } y^2 \cdot y^2=y^4)}\\[4pt] (b^3+5)(b^3-9)&=b^6-4b^3-45\quad{\color{Red}(\text{notice } b^3 \cdot b^3=b^6)} \end{align*}\]

Although we will not focus on factoring trinomials with more than one variable in this course, this pattern also holds for true for product of binomials with multiple variables:

\[\begin{align*}(kl+5)(kl-10)&=k^2l^2-5kl-50 \quad{\color{Red}(\text{notice that } kl \cdot kl=k^2l^2)} \\[4pt] (m^4n-3)(m^4n-7)&=m^8n^2-10m^4n+21\quad{\color{Red}(\text{notice that } m^4n \cdot m^4n=m^8n^2)}\end{align*}\]

There is also a pattern with the signs of terms of the binomials and the signs of the terms of the trinomial. Let's look back at some earlier examples to investigate this pattern:

\[\begin{align*}(x+2)(x+3)&=x^2+5x+6\\[4pt] (x-2)(x-3)&=x^2-5x+6 \end{align*}\]

Each of these examples contain a product of binomials that have the same signs, both "+" or both "-". Notice that the last term of the trinomial written in descending order is positive in both cases and that the middle term is the same sign as the signs of the binomials.

Next, let's look at the pattern of the following:

\[\begin{align*}(x+2)(x-3)&=x^2-x-6\\[4pt] (x-2)(x+3)&=x^2+x-6 \end{align*}\]

Each of these examples contain a product of binomials that have the different signs, one contains a "+" sign and the other contains a "-" sign. The last term of the trinomial written in descending order is negative in both examples.

The next task is to make sure that we can properly identify the coefficients of a trinomial.

Compare \(x^2−8x−9\) with the form \(ax^2 +bx+c\) and identify the coefficients \(a\), \(b\), and \(c\).

**Solution**

Align the trinomial \(x^2−8x−9\) with the standard form \(ax^2 +bx+c\), then compare coefficients. Note that the understood coefficient of \(x^2\) is \(1\).

\[\begin{array}{l}{a x^{2}+b x+c} \\ {{\color {Red}1} x^{2}-8 x-9}\end{array} \nonumber \]

We see that \(a = 1\), \(b = −8\), and \(c = −9\). Because the leading coefficient is \(1\), this is the type of trinomial that we will learn how to factor in this section.

Compare \(2x^2 +5x−3\) with the form \(ax^2 + bx + c\) and identify the coefficients \(a\), \(b\), and \(c\).

**Answer**-
\(a = 2\), \(b = 5\), \(c =−3\)

Compare \(−40 + 6x^2 − x\) with the form \(ax^2 + bx + c\) and identify the coefficients \(a\), \(b\), and \(c\).

**Solution**

First, arrange \(−40+ 6x^2 −x\) in descending powers of \(x\), then align it with the standard form \(ax^2 +bx+c\) and compare coefficients. Note that the understood coefficient of \(x\) is \(−1\).

\[\begin{array}{l}{a x^{2}+b x+c} \\ {6 x^{2}-{\color {Red} 1} x-40}\end{array} \nonumber \]

We see that \(a = 6\), \(b = −1\), and \(c = −40\). Because the leading coefficient is \(6\), we will have to wait until we learn about factoring \(ax^2 + bx + c\) when \(a \neq 1\) in section 6.4 before learning how to factor this trinomial.

Compare \(3x +9−7x^2\) with the form \(ax^2 + bx + c\) and identify the coefficients \(a\), \(b\), and \(c\).

**Answer**-
\(a = −7\), \(b = 3\), \(c =9\)

## The \(ac\)-Method

We are now going to introduce a technique called the \(ac\)-method (or \(ac\)-test) for factoring trinomials of the form \(ax^2 +bx+c\) when \(a = 1\). In the next section we will learn how to factor \(ax^2 + bx + c\) when \(a \neq 1\). We will see that this method can also be employed when \(a \neq 1\), with one minor exception. But for the remainder of this section, we focus strictly on trinomials whose leading coefficient is \(1\).

Let’s begin by finding the following product:

\[\begin{aligned}(x+12)(x-4) &=x(x-4)+12(x-4) \quad \color {Red} \text { Apply the distributive property. } \\ &=x^{2}-4 x+12 x-48 \quad \color {Red } \text { Distribute again. } \\ &=x^{2}+8 x-48 \quad \color {Red} \text { Simplify. }\end{aligned} \nonumber \]

Now, can we reverse the process? That is, can we start with \(x^2 +8x−48\) and place it in its original factored form \((x + 12)(x−4)\)? The answer is yes, if we apply the following procedure.

Compare the given polynomial with the standard form \(ax^2 + bx + c\), determine the coefficients \(a\), \(b\), and \(c\), then proceed as follows:

- Multiply the coefficients \(a\) and \(c\) and determine their product \(ac\). List all the integer pairs whose product equals \(ac\).
- Circle the pair in the list produced in step 1 whose sum equals \(b\), the coefficient of the middle term of \(ax^2 + bx + c\).
- Replace the middle term \(bx\) with a sum of like terms using the circled pair from step 2.
- Factor by grouping.
- Check the result using the
**FOIL**shortcut.

Let’s follow the steps of the \(ac\)-method to factor \(x^2 +8x−48\).

Factor: \(x^2 +8x−48\).

**Solution**

The trinomial contains an \(x^2\) term, an \(x\) term, and a constant. The polynomial is in descending order and the last term is negative which tells us the binomials will have different signs. So, the factorization will have the form \((x+\text{constant})(x-\text{constant})\).

binomials Compare \(x^2+8x−48\) with \(ax^2+bx+c\) and identify \(a = 1\), \(b = 8\), and \(c = −48\). Note that the leading coefficient is \(a = 1\). Calculate \(ac\). Note that \(ac = (1)(−48)\), so \(ac =−48\). List all integer pairs whose product is \(ac = −48\).

\[\begin{array}{ll}{1,-48} & {-1,48} \\ {2,-24} & {-2,24} \\ {3,-16} & {-3,16} \\ {4,-12} & { -4,12} \\ {6,-8} & {-6,8}\end{array} \nonumber \]

Circle the ordered pair whose sum is \(b = 8\).

\[\begin{array}{ll}{1,-48} & {-1,48} \\ {2,-24} & {-2,24} \\ {3,-16} & {-3,16} \\ {4,-12} & {\color {Red}-4,12} \\ {6,-8} & {-6,8}\end{array} \nonumber \]

Replace the middle term \(8x\) with a sum of like terms using the circled pair whose sum is \(8\).

\[x^2{\color {Red}+8x}-48 = x^2{\color {Red}-4x + 12x}-48 \nonumber \]

Factor by grouping.

\[\begin{aligned} x^{2}+8 x-48 &=x(x-4)+12(x-4) \\ &=(x+12)(x-4) \end{aligned} \nonumber \]

Use the **FOIL** shortcut to mentally check your answer. To determine the product \((x + 12)(x−4)\), use these steps:

- Multiply the terms in the “First” positions: \(x^2\).
- Multiply the terms in the “Outer” and “Inner” positions and combine the results mentally: \(−4x + 12x =8x\).
- Multiply the terms in the “Last” positions: \(−48\).

That is:

\[(x+12)(x-4) = \begin{array}{ccccccc} {\color {Red}F} & & {\color {Red}O} & & {\color {Red}I} & & {\color {Red}L}\\ x^2&-&4x&+&12x&-&48 \end{array} \nonumber \]

Combining like terms, \((x + 12)(x−4) = x^2 +8x−48\), which is the original trinomial, so our solution checks. Note that if you combine the “Outer” and “Inner” products mentally, the check goes even faster.

Factor: \(m^2 + 11m + 28\)

**Answer**-
\((m+4)(m+7)\)

Some readers might ask “Is it a coincidence that the circled pair \(\color {Red} −4,12\) seemed to ‘drop in place’ in the resulting factorization \((x+12)(x−4)\)?” Before we answer that question, let’s try another example.

Factor: \(y^4 −9y^2−36\).

**Solution**

Notice that there is a constant term and if we take the variable of the middle term and multiply it by itself that we get y^4. Also, the last term is negative.This indicates that \(y^4 −9y^2−36\) factors into the form \((y^2+ \text{constant})((y^2-\text{constant})\).

Comparing \(y^4−9y^2−36\) to the form \(ax^2+bx+c\), \(a = 1\), \(b = −9\), and \(c =−36\). Calculate \(ac = (1)(−36)\), so \(ac = −36\).

At this point, some readers might ask “What if I start listing the ordered pairs and I see the pair I need? Do I need to continue listing the remaining pairs?

The answer is “No.” In this case, we start listing the integer pairs whose product is \(ac = −36\), but are mindful that we need an integer pair whose sum is \(b =−9\). The integer pair \(3\) and \(−12\) has a product equaling \(ac =−36\) and a sum equaling \(b =−9\).

\[\begin{array}{r}{1,-36} \\ {2,-18} \\ {\color {Red}3,-12} \end{array} \nonumber \]

Note how we ceased listing ordered pairs the moment we found the pair we needed. Next, replace the middle term \(−9x\) with a sum of like terms using the circled pair.

\[y^4{\color {Red}-9y^2}-36 = y^4{\color {Red}+3y^2-12y^2}-36 \nonumber \]

Factor by grouping.

\[\begin{aligned} y^{4}{\color {Red}-9 y^2}-36 &=y^2(y^2+3)-12(y^2+3) \\ &=(y^2-12)(y^2+3) \end{aligned} \nonumber \]

It is left to the reader to se the **FOIL** shortcut to check your answer.

Factor: \(b^6 + 10b^3−24\).

**Answer**-
\(( b^3 + 12)(b^3−2)\)

## Speeding Things Up a Bit

Readers might again ask “Is it a coincidence that the circled pair \(\color {Red} 3,−12\) seemed to ‘drop in place’ in the resulting factoriztion \((x−12)(x + 3)\)?” The answer is “No,” it is not a coincidence. Provided the leading coefficient of the trinomial \(ax^2 +bx+c\) is \(a = 1\), you can always “drop in place” the circled pair in order to arrive at the final factorization, skipping the factoring by grouping.

Some readers might also be asking “Do I really have to list any of those ordered pairs if I already recognize the pair I need?” The answer is “No!” If you see the pair you need, drop it in place.

Factor: \(x^2 −5x−24\).

**Solution**

Compare \(x^2−5x−24\) with \(ax^2+bx+c\) and note that \(a = 1\), \(b =−5\), and \(c =−24\). Calculate \(ac = (1)(−24)\), so \(ac = −24\). Now can you think of an integer pair whose product is \(ac = −24\) and whose sum is \(b = −5\)? For some, the required pair just pops into their head: \(−8\) and \(3\). The product of these two integers is \(−24\) and their sum is \(−5\).“Drop” this pair in place and you are done.

\[x^2 −5x−24 = (x−8)(x + 3) \nonumber \]

Use the **FOIL** shortcut to check your answer.

\[(x-8)(x+3) = \begin{array}{ccccccc} {\color {Red}F} & & {\color {Red}O} & & {\color {Red}I} & & {\color {Red}L}\\ x^2&+&3x&-&8x&-&24 \end{array} \nonumber\]

Combining like terms, \((x−8)(x + 3) =x^2 −5x−24\), the original trinomial. Our solution checks.

Factor: \(a^4 −12a^2 + 35\)

**Answer**-
\((a^2−7)(a^2−5)\)

The “Drop in Place” technique of Example \(\PageIndex{5}\) allows us to revise the \(ac\) method a bit.

Compare the given polynomial with the standard form \(ax^2 + bx + c\), determine the coefficients \(a\), \(b\), and \(c\), then determine a pair of integers whose product equals \(ac\) and whose sum equals \(b\). You then have two options:

- Write the middle term as a product of like terms using the ordered pair whose product is \(ac\) and whose sum is \(b\). Complete the factorization process by factoring by grouping.
- Simply “drop in place” the ordered pair whose product is \(ac\) and whose sum is \(b\) to complete the factorization process.

*Note: This "drop in place" method only works when \(a=1\). If \(a=-1\), then factor "-1" out of the polynomial first before identifying the coefficients \(a\), \(b\), and \(c\).*

Readers are strongly encouraged to check their factorization by determining the product using the **FOIL** method. If this produces the original trinomial, the factorization is correct.

## Nonlinear Equations Revisited

The ability to factor trinomials of the form \(ax^2 +bx+c\), where \(a = 1\), increases the number of nonlinear equations we are now able to solve.

Solve the equation \(x^2 =2 x + 3\) both algebraically and graphically, then compare your answers.

**Solution**

Because there is a power of \(x\) larger than one, the equation is nonlinear. Make one side zero.

\[\begin{aligned} x^{2} &=2 x+3 \quad \color {Red} \text { Original equation. } \\ x^{2}-2 x &=3 \quad \color {Red} \text { Subtract } 2 x \text { from both sides. } \\ x^{2}-2 x-3 &=0 \quad \color {Red} \text { Subtract } 3 \text { from both sides. }\end{aligned} \nonumber \]

Compare \(x^2−2x−3\) with \(ax^2+bx+c\) and note that \(a = 1\), \(b =−2\) and \(c =−3\). We need an integer pair whose product is \(ac = −3\) and whose sum is \(b = −2\). The integer pair \(1\) and \(−3\) comes to mind. “Drop” these in place to factor.

\[(x + 1)(x-3) = 0 \quad \color {Red} \text {Factor.} \nonumber \]

We have a product that equals zero. Use the zero product property to complete the solution.

\[\begin{aligned} x+1 &=0 \\ x &=-1 \end{aligned} \nonumber \]

or

\[\begin{array}{r}{x-3=0} \\ {x=3}\end{array} \nonumber \]

Thus, the solutions of \(x^2 =2x + 3\) are \(x = −1\) and \(x = 3\).

**Checking Solution Using a Graphing Calculator:**

Load each side of the equation \(x^2 =2 x + 3\) into the **Y= menu** of your graphing calculator, \(y = x^2\) in \(\mathbb {Y1}\), \(y =2x + 3\) in \(\mathbb {Y2}\) (see Figure \(\PageIndex{1}\)). Select **6:ZStandard** from the **ZOOM** menu to produce the image at the right in Figure \(\PageIndex{1}\).

One of the intersection points is visible on the left, but the second point of intersection is very near the top of the screen at the right (see Figure \(\PageIndex{1}\)). Let’s extend the top of the screen a bit. Press the **WINDOW** button and make adjustments to \(\mathbb{Ymin}\) and \(\mathbb{Ymax}\) (see Figure \(\PageIndex{2}\)), then press the **GRAPH** button to adopt the changes.

Note that both points of intersection are now visible in the viewing window (see Figure \(\PageIndex{2}\)). To find the coordinates of the points of intersection, select **5:intersect** from the **CALC** menu. Press the **ENTER** key to accept the “First curve,” press **ENTER** again to accept the “Second curve,” then press **ENTER** again to accept the current position of the cursor as your guess. The result is shown in the image on the left in Figure \(\PageIndex{3}\). Repeat the process to find the second point of intersection, only when it comes time to enter your “Guess,” use the right-arrow key to move the cursor closer to the second point of intersection than the first.

Solve the equation \(x^2 = −3x + 4\) both algebraically and graphically, then compare your answers.

**Answer**-
\(-4\), \(1\)

Solve the equation \(x^2 −4x−96 = 0\) both algebraically and graphically, then compare your answers.

**Solution**

Because there is a power of \(x\) larger than one, the equation \(x^2 − 4x−96 = 0\) is nonlinear. We already have one side zero, so we can proceed with the factoring. Begin listing integer pairs whose product is \(ac = −96\), mindful of the fact that we need a pair whose sum is \(b = −4\).

\[\begin{array}{l}{1,-96} \\ {2,-48} \\ {3,-32} \\ {4,-24} \\ {6,-16} \\ {\color {Red}8,-12} \\ \end{array} \nonumber \]

Note that we stopped the listing process as soon as we encountered a pair whose sum was \(b = −4\). “Drop” this pair in place to factor the trinomial.

\[\begin{aligned}x^{2}-4 x-96 &=0 \quad \color {Red} \text { Original equation. } \\ (x+8)(x-12) &=0 \quad \color {Red} \text { Factor. }\end{aligned} \nonumber \]

We have a product that equals zero. Use the zero product property to complete the solution.

\[\begin{aligned} x+8 &=0 \\ x &=-8 \end{aligned} \nonumber \]

or

\[\begin{aligned} x-12 &=0 \\ x &=12 \end{aligned} \nonumber \]

Thus, the solutions of \(x^2 −4x−96 = 0\) are \(x =−8\) and \(x = 12\).

**Graphical solution:**

Load the equation \(y = x^2 −4x−96\) in \(\mathbb{Y1}\) in the **Y= menu** of your graphing calculator (see Figure \(\PageIndex{5}\)). Select **6:ZStandard** from the **ZOOM** menu to produce the image at the right in Figure \(\PageIndex{5}\).

When the degree of a polynomial is two, we’re used to seeing some sort of parabola. In Figure \(\PageIndex{5}\), we saw the graph go down and off the screen, but we did not see it turn and come back up. Let’s adjust the **WINDOW** parameters so that the vertex (turning point) of the parabola and both \(x\)-intercepts are visible in the viewing window. After some experimentation, the settings shown in Figure \(\PageIndex{6}\) reveal the vertex and the \(x\)-intercepts. Press the **GRAPH** button to produce the image at the right in Figure \(\PageIndex{6}\).

Note that both \(x\)-intercepts of the parabola are now visible in the viewing window (see Figure \(\PageIndex{6}\)). To find the coordinates of the \(x\)-intercepts, select **2:zero** from the **CALC** menu. Use the left- and right-arrow keys to move the cursor to the left of the first \(x\)-intercept, then press **ENTER** to mark the “Left bound.” Next, move the cursor to the right of the first \(x\)-intercept, then press **ENTER** to mark the “Right bound.” Press **ENTER** to accept the current position of the cursor as your “Guess.” The result is shown in the image on the left in Figure \(\PageIndex{7}\). Repeat the process to find the coordinates of the second \(x\)-intercept. The result is show in the image on the right in Figure \(\PageIndex{7}\).

Solve the equation \(x^2 −21x + 90 = 0\) both algebraically and graphically, then compare your answers.

**Answer**-
\(6\), \(15\)