# Modeling, Functions, and Graphs

Consider the trinomial
\begin{equation*} x^2 + 10x + 16 \end{equation*}
Can we find two binomial factors,
\begin{equation*} (x + a)(x + b) \end{equation*}
whose product is the given trinomial? The product of the binomials is
\begin{equation*} (x + a)(x + b) = x^2 + (a + b)x + ab \end{equation*}
Thus, we are looking for two numbers, $$a$$ and $$b\text{,}$$ that satisfy
\begin{equation*} (x + a)(x + b) = x^2 + (\alert{a + b})x + \blert{ab}= x^2 + \alert{10}x + \blert{16} \end{equation*}
By comparing the coefficients of the terms in the two trinomials, we see that $$\alert{a + b = 10}$$ and $$\blert{ab = 16}\text{.}$$ That is, the sum of the two numbers is the coefficient of the linear term, $$10\text{,}$$ and their product is the constant term, $$16\text{.}$$
To find the numbers, we list all the possible integer factorizations of $$16\text{:}$$
\begin{equation*} 1 \cdot 16, ~~~2 \cdot 8, ~~~\text{ and }~~~ 4 \cdot 4 \end{equation*}
We see that only one combination gives the correct linear term: $$8$$ and $$2\text{.}$$ These are the numbers $$a$$ and $$b\text{,}$$ so
\begin{equation*} x^2 + 10x + 16 = (x + 8)(x + 2) \end{equation*}
In Example A.57 we factor quadratic trinomials in which one or more of the coefficients is negative.

### ExampleA.57.

Factor.
1. $$\displaystyle x^2-7x+12$$
2. $$\displaystyle x^2-x-12$$
Solution.
1. Find two numbers whose product is $$12$$ and whose sum is $$-7\text{.}$$ Because the product is positive and the sum is negative, the two numbers must both be negative. The possible factors of $$12$$ are $$-1$$ and $$-12\text{,}$$ $$-2$$ and $$-6\text{,}$$ or $$-3$$ and $$-4\text{.}$$ Only $$-4$$ and $$-3$$ have the correct sum, $$-7\text{.}$$ Hence,
\begin{equation*} x^2 - 7x + 12 = (x - 4) (x - 3) \end{equation*}
2. Find two numbers whose product is $$-12$$ and whose sum is $$-1\text{.}$$ Because the product is negative, the two numbers must be of opposite sign and their sum must be $$-1\text{.}$$ By listing the possible factors of $$-12\text{,}$$ we find that the two numbers are $$-4$$ and $$3\text{.}$$ Hence,
\begin{equation*} x^2 - x - 12 = (x - 4) (x + 3) \end{equation*}
If the coefficient of the quadratic term is not $$1\text{,}$$ we must also consider its factors.

### ExampleA.58.

Factor $$~8x^2 - 9 - 21x$$
Solution.
1. Write the trinomial in decreasing powers of $$x\text{.}$$
\begin{equation*} 8x^2-21x-9 \end{equation*}
2. List the possible factors for the quadratic term.
\begin{gather*} (8x\hphantom{00000})(x\hphantom{000000})\\ (4x\hphantom{00000})(2x\hphantom{00000}) \end{gather*}
3. Consider possible factors for the constant term: $$9$$ may be factored as $$9\cdot 1$$ or as $$3\cdot 3\text{.}$$ Form all possible pairs of binomial factor using these factorizations.
4. Select the combinations of the products $$\Large{①}$$ and $$\Large{②}$$ whose sum or difference could be the linear term, $$-21x\text{.}$$
\begin{equation*} (8x \hphantom{000} 3) (x \hphantom{000} 3) \end{equation*}
5. Insert the proper signs:
\begin{equation*} (8x + 3) (x - 3) \end{equation*}
With practice, you can usually factor trinomials of the form $$Ax^2 + Bx + C$$ mentally. The following observations may help.
1. If $$A\text{,}$$ $$B$$ and $$C$$ are all positive, both signs in the factored form are positive. For example, as a first step in factoring $$6x^2 + 11x + 4\text{,}$$ we could write
\begin{equation*} (\hphantom{000} + \hphantom{000} ) (\hphantom{000} + \hphantom{000} ) \end{equation*}
2. If $$A$$ and $$C$$ are positive and $$B$$ is negative, both signs in the factored form are negative. Thus as a first step in factoring $$6x^2 - 11x + 4\text{,}$$ we could write
\begin{equation*} (\hphantom{000} - \hphantom{000} ) (\hphantom{000} - \hphantom{000} ) \end{equation*}
3. If $$C$$ is negative, the signs in the factored form are opposite. Thus as a first step in factoring $$6x^2 - 5x - 4\text{,}$$ we could write
\begin{equation*} (\hphantom{000} + \hphantom{000} ) (\hphantom{000} - \hphantom{000} ) ~~\text{ or }~~ (\hphantom{000} - \hphantom{000} ) (\hphantom{000} + \hphantom{000} ) \end{equation*}

### ExampleA.59.

1. \displaystyle \begin{aligned}[t] 6x^2 + 5x + 1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} + \hphantom{00} )\\ \amp = (3x + 1) (2x + 1) \end{aligned}
2. \displaystyle \begin{aligned}[t] 6x^2 - 5x + 1 \amp = (\hphantom{00} - \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x - 1) (2x - 1) \end{aligned}
3. \displaystyle \begin{aligned}[t] 6x^2 - x - 1 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x + 1) (2x - 1) \end{aligned}
4. \displaystyle \begin{aligned}[t] 6x^2 - xy - y^2 \amp = (\hphantom{00} + \hphantom{00} ) (\hphantom{00} - \hphantom{00} )\\ \amp = (3x + y) (2x - y) \end{aligned}

### SubsectionSpecial Products and Factors

The products below are special cases of the multiplication of binomials. They occur so often that you should learn to recognize them on sight.

#### Special Products.

1. $$\displaystyle (a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2$$
2. $$\displaystyle (a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2$$
3. $$\displaystyle (a + b)(a-b) = a^2 - b^2$$

#### CautionA.60.

Notice that in (I) $$~~(a + b)^2 \ne a^2 + b^2~~\text{,}$$ and that in (II) $$~~(a - b)^2\ne a^2 - b^2~~\text{.}$$ For example,
\begin{align*} (x + 4)^2 \amp\ne x^2 + 16, \amp\amp\text{ instead } \amp (x + 4)^2 \amp = x^2 + 8x + 16\\ (t-5)^2 \amp\ne t^2 - 16, \amp\amp\text{ instead } \amp (t -5)^2 \amp= t^2 -10t + 25 \end{align*}

#### ExampleA.61.

1. \displaystyle \begin{aligned}[t] 3(x + 4)^2 \amp = 3(x^2 + 2 \cdot 4x + 4^2) \\ \amp = 3x^2 + 24x + 48 \end{aligned}
2. \displaystyle \begin{aligned}[t] (y + 5) (y - 5) \amp = y^2-5^2 \\ \amp = y^2-25 \end{aligned}
3. \displaystyle \begin{aligned}[t] (3x - 2y)^2 \amp = (3x)^2 - 2(3x)(2y) + (2y)^2 \\ \amp = 9x^2-12xy+4y^2 \end{aligned}
Each of the formulas for special products, when viewed from right to left, also represents a special case of factoring quadratic polynomials.

#### Special Factorizations.

1. $$\displaystyle a^2 + 2ab + b^2=(a + b)^2$$
2. $$\displaystyle a^2 - 2ab + b^2=(a - b)^2$$
3. $$\displaystyle a^2 - b^2=(a + b)(a-b)$$
4. $$\displaystyle a^2+b^2 ~~\text{ cannot be factored}$$
The trinomials in (I) and (II) are sometimes called perfect-square trinomials because they are squares of binomials. Note that the sum of two squares, $$a^2 + b^2\text{,}$$ cannot be factored.

#### ExampleA.62.

Factor.
1. $$\displaystyle x^2 + 8x + 16$$
2. $$\displaystyle y^2 -10y + 25$$
3. $$\displaystyle 4a^2 - 12ab + 9b^2$$
4. $$\displaystyle 25m^2n^2 + 20mn + 4$$
Solution.
1. Because $$16$$ is equal to $$4^2$$ and $$8$$ is equal to $$2\cdot 4\text{,}$$
\begin{align*} x^2 + 8x + 16 \amp = x^2 - 2 \cdot 4x + 4^2\\ \amp = (x + 4)^2 \end{align*}
2. Because $$25=5^2$$ and $$10=2\cdot 5\text{,}$$
\begin{align*} y^2 - 10y + 25 \amp = y^2 - 2 \cdot 5y + 5^2\\ \amp = (y-5)^2 \end{align*}
3. Because $$4a^2=(2a)^2\text{,}$$ $$9b^2=(3b)^2\text{,}$$ and $$2ab=2(2a)(3b)\text{,}$$
\begin{align*} 4a^2 - 12ab + 9b^2 \amp = (2a)^2 - 2(2a)(3b) + (3b)^2\\ \amp = (2a-3b)^2 \end{align*}
4. Because $$25m^2n^2=(5mn)^2\text{,}$$ $$4=2^2\text{,}$$ and $$20mn=2(5mn)(2)\text{,}$$
\begin{align*} 25m^2n^2 + 20mn + 4 \amp = (5mn)^2 + 2(5mn)(2) + 2^2\\ \amp = (5mn+2)^2 \end{align*}
Binomials of the form $$a^2 - b^2$$ are often called the difference of two squares.

#### ExampleA.63.

Factor if possible.
1. $$\displaystyle x^2 - 81$$
2. $$\displaystyle 4x^2 - 9y^2$$
3. $$\displaystyle x^2 + 81$$
Solution.
1. The expression $$x^2 - 81$$ is the difference of two squares, $$x^2 - 9^2\text{,}$$ and thus can be factored according to Special Factorization (III) above.
\begin{align*} x^2-81 \amp = x^2-9^2\\ \amp = (x+9)(x-9) \end{align*}
2. Because $$4x^2 - 9y^2$$ can be written as $$(2x)^2 -(3y)^2\text{,}$$
\begin{align*} 4x^2 - 9y^2 \amp =(2x^2)-(3y)^2 \\ \amp =(2x+3y)(2x-3y) \end{align*}
3. The expression $$x^2 + 81\text{,}$$ or $$x^2 + 0x + 81\text{,}$$ is not factorable, because no two real numbers have a product of $$81$$ and a sum of $$0\text{.}$$

#### CautionA.64.

$$x^2 + 81\ne (x + 9) (x + 9)\text{,}$$ which you can verify by multiplying
\begin{equation*} (x + 9) (x + 9) = x^2 + 18x + 8 \end{equation*}
The factors $$x + 9$$ and $$x - 9$$ in Example A.63a are called conjugates of each other. In general, any binomials of the form $$a+b$$ and $$a-b$$ are called a conjugate pair.

### SubsectionSection Summary

#### SubsubsectionVocabulary

Look up the definitions of new terms in the Glossary.
• Perfect-square trinomial
• Difference of squares
• Conjugate

#### SubsubsectionSKILLS

Practice each skill in the exercises listed.
2. Expand special products: #37–48
3. Factor special quadratic expressions: #49–68

### ExercisesExercises A.8

#### Exercise Group.

For Problems 1-36, factor completely.
##### 1.
$$x^2+5x+6$$
##### 2.
$$x^2+5x+4$$
##### 3.
$$y^2-7y+12$$
##### 4.
$$y^2-7y+10$$
##### 5.
$$x^2-6-x$$
##### 6.
$$x^2-15-2x$$
##### 7.
$$2x^2+3x-2$$
##### 8.
$$3x^2-7x+2$$
##### 9.
$$7x+4x^2-2$$
##### 10.
$$1-5x+6x^2$$
##### 11.
$$9y^2-21y-8$$
##### 12.
$$10y^2-3y-18$$
##### 13.
$$10u^2-3-u$$
##### 14.
$$8u^2-3+5u$$
##### 15.
$$21x^2-43x-14$$
##### 16.
$$24x^2-29x+5$$
##### 17.
$$5a+72a^2-25$$
##### 18.
$$-30a+72a^2-25$$
##### 19.
$$12-53x+30x^2$$
##### 20.
$$39x+80x^2-20$$
##### 21.
$$-30t-44+54t^2$$
##### 22.
$$48t^2-122t+39$$
##### 23.
$$3x^2-7ax+2a^2$$
##### 24.
$$9x^2+9ax-10a^2$$
##### 25.
$$15x^2-4xy-4y^2$$
##### 26.
$$12x^2+7xy-12y^2$$
##### 27.
$$18u^2+20v^2-39uv$$
##### 28.
$$24u^2-20v^2+17uv$$
##### 29.
$$12a^2-14b^2-13ab$$
##### 30.
$$24a^2-15b^2-2ab$$
##### 31.
$$10a^2b^2-19ab+6$$
##### 32.
$$12a^2b^2-ab-20$$
##### 33.
$$56x^2y^2-2xy-4$$
##### 34.
$$54x^2y^2+3xy-2$$
##### 35.
$$22a^2z^2-21-19az$$
##### 36.
$$26a^2z^2-24+23az$$

#### Exercise Group.

For Problems 37-48, write the expression as a polynomial and simplify.
##### 37.
$$(x+3)^2$$
##### 38.
$$(y-4)^2$$
##### 39.
$$(2y-5)^2$$
##### 40.
$$(3x+2)^2$$
##### 41.
$$(x+3)(x-3)$$
##### 42.
$$(x-7)(x+7)$$
##### 43.
$$(3t-4s)(3t+4s)$$
##### 44.
$$(2x+a)(2x-a)$$
##### 45.
$$(5a-2)(5a-2)$$
##### 46.
$$(4u+5v)(4u+5v)$$
##### 47.
$$(8xz+3)(8xz+3)$$
##### 48.
$$(7yz-2)(7yz-2)$$

#### Exercise Group.

For Problems 49-68, factor completely.
##### 49.
$$x^2-25$$
##### 50.
$$x^2-36$$
##### 51.
$$x^2-24x+144$$
##### 52.
$$x^2+26x+169$$
##### 53.
$$x^2-4y^2$$
##### 54.
$$9x^2-y^2$$
##### 55.
$$4x^2+12x+9$$
##### 56.
$$4y^2+4y+1$$
##### 57.
$$9u^2-30uv+25v^2$$
##### 58.
$$16s^2-56st+49t^2$$
##### 59.
$$4a^2-25b^2$$
##### 60.
$$16a^2-9b^2$$
##### 61.
$$x^2y^2-81$$
##### 62.
$$x^2y^2-64$$
##### 63.
$$9x^2y^2+6xy+1$$
##### 64.
$$4x^2y^2+12xy+9$$
##### 65.
$$16^2y^2-1$$
##### 66.
$$64x^2y^2-1$$
##### 67.
$$(x+2)^2-y^2$$
##### 68.
$$x^2-(y-3)^2$$