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2.7 — The x Factor
In Section 2.6 you learned to express the area of a rectangle as a product or as an expanded polynomial. Today we extend that method to rectangles with dimensions smaller as well as larger than the original side length. This makes the signs in quadratic trinomials predictable and makes factoring easier.
1. In class: A Solidify Understanding Task
Multiply binomials by thinking about rectangular areas
Begin with a square of side length \(x\). If one side stays at \(x\) and the other becomes \(x+2\), the area is \(x(x+2)=x^2+2x\). If the other side becomes \(x-2\), the area is \(x(x-2)=x^2-2x\). We require \(x>2\) for the second expression to represent an ordinary physical rectangle.
A rectangle has dimensions \(x-2\) and \(x+3\). Write its area in two equivalent ways.
\[\begin{aligned}A&=(x-2)(x+3)\\&=x^2+3x-2x-6\\&=x^2+x-6.\end{aligned}\]The four products are \(x\cdot x\), \(x\cdot3\), \((-2)\cdot x\), and \((-2)\cdot3\). The rectangle is physically meaningful when \(x>2\). The algebraic identity holds for every real \(x\).
In general, multiplying \((x+m)(x+n)\) gives four pieces:
The coefficient of \(x\) is the sum of the side changes; the constant term is their product. This remains true for negative numbers.
Suppose the dimensions are \(x-3\) and \(x-4\). Then
\[\begin{aligned}(x-3)(x-4)&=x^2-4x-3x+12\\&=x^2-7x+12.\end{aligned}\]The negative side changes add to \(-7\), but their product is positive \(12\). Physically require \(x>4\).
Predict the signs of a trinomial
For a monic quadratic \(x^2+bx+c\), the word monic means the coefficient of \(x^2\) is 1. If the expression comes from \((x+m)(x+n)\), then \(b=m+n\) and \(c=mn\).
| Signs of \(m,n\) | Sign of constant \(c=mn\) | Sign of middle coefficient \(b=m+n\) |
|---|---|---|
| Both positive | Positive | Positive |
| Both negative | Positive | Negative |
| Opposite signs | Negative | Determined by which magnitude is greater; possibly zero |
Compare these products:
\[\begin{aligned}(x+6)(x-2)&=x^2+4x-12,\\(x+2)(x-6)&=x^2-4x-12.\end{aligned}\]In each, \(c=-12\). The sign of \(b\) changes because the larger-magnitude side change switches from positive to negative.
Equal increases and decreases: the difference of squares
Suppose one side of an \(x\)-by-\(x\) square grows by 4, and the other shrinks by 4. The rectangle's new area is not unchanged:
The two rectangular strip terms cancel, but the corner correction \(-16\) remains. For positive dimensions, \(x>4\).
To factor a difference of squares, reverse this identity: \(u^2-v^2=(u-v)(u+v)\). It works because the two cross terms cancel.
Check: \((x-9)(x+9)=x^2+9x-9x-81=x^2-81\).
Recognize also the special square patterns from earlier lessons: \(x^2+10x+25=(x+5)^2\) and \(x^2-10x+25=(x-5)^2\). These are different from a difference of two squares.
Reverse the multiplication: factor a trinomial
Factoring means rewriting an expanded expression as a product. For \(x^2+bx+c\), find two numbers \(m,n\) for which both conditions hold:
Then \(x^2+bx+c=(x+m)(x+n)\). In the assigned tasks, use integer pairs; check every answer by multiplying.
Factor \(x^2+5x-24\). Since the constant is negative, the two numbers have opposite signs. Consider the factor pairs of 24: \(1,24\), \(2,12\), \(3,8\), \(4,6\). The pair \(8\) and \(-3\) has product \(-24\) and sum \(5\).
\[x^2+5x-24=(x+8)(x-3).\]Verify: \(x^2-3x+8x-24=x^2+5x-24\).
Factor \(x^2-13x+36\). Since the product is positive and the sum is negative, both numbers must be negative. The factor pair \(4\) and \(9\) of 36 gives the needed sum 13, so use \(-4,-9\):
\[x^2-13x+36=(x-4)(x-9).\]Factor \(x^2-12x+36\). The numbers \(-6,-6\) multiply to \(36\) and sum to \(-12\):
\[x^2-12x+36=(x-6)^2.\]This is a perfect-square trinomial; always check for matching factors.
Try \(x^2+4x+7\). Integer factor pairs of 7 are \(1,7\) and \(-1,-7\), with sums \(8\) and \(-8\), not 4. Therefore it cannot be written as \((x+m)(x+n)\) with integer \(m,n\). That is a statement about the specified factoring method, not a claim that every quadratic factors over the integers.
2. Ready review — Finding numbers between two real numbers
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The density property of real numbers says that whenever \(a<b\), there is another real number strictly between them. In fact, there are infinitely many. One is the midpoint \(\frac{a+b}{2}\). By repeatedly taking midpoints, we can find three or more distinct numbers between any two values.
Find three values strictly between \(2\frac14=2.25\) and \(2\frac34=2.75\).
Choose \(2.3\), \(2.5\), and \(2.7\). All obey \(2.25<2.3<2.5<2.7<2.75\).
Find three values between \(-3.2\) and \(-2.7\). In increasing order, use \(-3.1\), \(-3.0\), and \(-2.9\). Remember: \(-3.1\) is greater than \(-3.2\).
Find three numbers between \(\frac13\) and \(\frac12\). Convert to a common denominator 60: \(\frac13=\frac{20}{60}\) and \(\frac12=\frac{30}{60}\). Choose \(\frac{21}{60},\frac{25}{60},\frac{29}{60}\); all are strictly between the endpoints.
Square roots and irrational endpoints
Square-root endpoints can be handled with careful decimal bounds or with squares. Since \(1.7^2=2.89<3\) and \(1.8^2=3.24>3\), \(1.7<\sqrt3<1.8\). Likewise \(2.2^2=4.84<5\) and \(2.3^2=5.29>5\), so \(2.2<\sqrt5<2.3\).
Find three numbers between \(\sqrt3\approx1.732\) and \(\sqrt5\approx2.236\). The choices \(1.8,2,2.2\) satisfy \(\sqrt3<1.8<2<2.2<\sqrt5\).
Find three numbers between 4 and \(\sqrt{23}\approx4.796\). Take \(4.2,4.4,4.6\). To justify the largest choice without relying solely on a calculator, \(4.6^2=21.16<23\), so \(4.6<\sqrt{23}\).
Between \(\sqrt{\frac14}=\frac12=0.5\) and \(\sqrt{\frac49}=\frac23\approx0.667\), pick \(0.55,0.60,0.65\). All lie strictly inside the interval.
The endpoints themselves are excluded. Use any three distinct real numbers strictly greater than the smaller endpoint and strictly less than the larger endpoint. The answers need not share the same form as the given numbers: decimals are acceptable between fractions or square roots, provided the comparisons are justified.
Check your work — Ready 1–8 answer key
- Examples: \(6,6.1,6.2\).
- Examples: \(-2,-1.9,-1.8\).
- Examples: \(0.3,0.4,0.5\).
- Examples: \(1.8,1.9,2\).
- Examples: \(4.1,4.2,4.3\).
- Examples: \(-9,-8.9,-8.8\).
- Examples: \(0.55,0.6,0.65\).
- Examples: \(3.61,3.62,3.63\).
3. Set practice — Recognize which factor pattern to use
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For the shared ideas used in these exercises, revisit the In class explanation.
The 20 Set exercises include many variations on \(x^2+bx+c\): positive or negative constants, positive or negative middle terms, repeated factors, and differences of squares. The same two-number test handles all of them.
| Pattern | Example | Why it works |
|---|---|---|
| Both additions | \(x^2+11x+30=(x+5)(x+6)\) | \(5+6=11\), \(5\cdot6=30\) |
| Both subtractions | \(x^2-9x+20=(x-4)(x-5)\) | \(-4-5=-9\), \((-4)(-5)=20\) |
| Mixed signs | \(x^2+2x-35=(x+7)(x-5)\) | \(7-5=2\), \(7(-5)=-35\) |
| Repeated binomial | \(x^2+8x+16=(x+4)^2\) | \(4+4=8\), \(4\cdot4=16\) |
| Difference of squares | \(x^2-49=(x-7)(x+7)\) | Middle terms cancel |
Factor \(x^2-19x+60\). The positive factor pairs of 60 are \((1,60),(2,30),(3,20),(4,15),(5,12),(6,10)\). The pair 4 and 15 sums to 19. Since the needed sum is negative, choose both negative:
\[x^2-19x+60=(x-4)(x-15).\]Check: \(-4-15=-19\) and \((-4)(-15)=60\).
Check your work — Set 9–28 answer key
- \(\left(x + 1\right) \left(x + 8\right)\).
- \(\left(x - 4\right) \left(x - 2\right)\).
- \(\left(x - 4\right) \left(x + 2\right)\).
- \(\left(x - 1\right) \left(x + 8\right)\).
- \(\left(x - 8\right) \left(x - 3\right)\).
- \(\left(x - 12\right) \left(x - 2\right)\).
- \(\left(x - 24\right) \left(x - 1\right)\).
- \(\left(x - 6\right) \left(x - 4\right)\).
- \(\left(x - 6\right) \left(x + 4\right)\).
- \(\left(x - 8\right) \left(x + 3\right)\).
- \(\left(x - 3\right) \left(x + 8\right)\).
- \(\left(x - 5\right)^{2}\).
- \(\left(x - 5\right) \left(x + 5\right)\).
- \(\left(x - 5\right) \left(x + 3\right)\).
- \(\left(x - 5\right) \left(x + 15\right)\).
- \(\left(x - 17\right) \left(x - 3\right)\).
- \(\left(x - 2\right) \left(x + 16\right)\).
- \(\left(x - 1\right) \left(x + 1\right)\).
- \(\left(x - 1\right)^{2}\).
- \(\left(x - 3\right) \left(x + 15\right)\).
4. Go review — Graphing parabolas and describing transformations
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The Go assignment reviews material from Sections 2.1–2.2. Start with the parent parabola \(y=x^2\), whose vertex is \((0,0)\). The axis of symmetry is the vertical line through the vertex. A horizontal shift changes the vertex's x-coordinate, a vertical shift changes its y-coordinate, and a negative sign outside the square reflects the graph across the x-axis.
Graph \(g(x)=x^2+2\). Its vertex is \((0,2)\), axis of symmetry is \(x=0\), and it opens upward. Plot seven points:
| \(x\) | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| \(g(x)\) | 11 | 6 | 3 | 2 | 3 | 6 | 11 |
Description: the graph of \(y=x^2\) moves up 2 units.
Graph \(h(x)=-(x+2)^2+3\). The vertex is \((-2,3)\) because \(x+2=0\) at \(x=-2\). The axis is \(x=-2\). The negative coefficient makes the graph open downward.
| \(x\) | −5 | −4 | −3 | −2 | −1 | 0 | 1 |
|---|---|---|---|---|---|---|---|
| \(h(x)\) | −6 | −1 | 2 | 3 | 2 | −1 | −6 |
Description: reflect \(y=x^2\) across the x-axis, shift 2 left, then shift 3 up.
Graph \(p(x)=(x-3)^2\). Vertex \((3,0)\), axis \(x=3\); opens up. Three pairs of points are \((2,1),(4,1)\), \((1,4),(5,4)\), and \((0,9),(6,9)\). Description: shift right by 3 units.
Check your work — Go 29–32 answer key
- Vertex \(\left( 0, \ 0\right)\); axis \(x=0\); points \(\left( -3, \ 9\right),\ \left( -2, \ 4\right),\ \left( -1, \ 1\right),\ \left( 1, \ 1\right),\ \left( 2, \ 4\right),\ \left( 3, \ 9\right)\). Plot not shown. Parent parabola; no transformation.
- Vertex \(\left( 0, \ -3\right)\); axis \(x=0\); points \(\left( -3, \ 6\right),\ \left( -2, \ 1\right),\ \left( -1, \ -2\right),\ \left( 1, \ -2\right),\ \left( 2, \ 1\right),\ \left( 3, \ 6\right)\). Plot not shown. Shift down 3.
- Vertex \(\left( 2, \ 0\right)\); axis \(x=2\); points \(\left( -1, \ 9\right),\ \left( 0, \ 4\right),\ \left( 1, \ 1\right),\ \left( 3, \ 1\right),\ \left( 4, \ 4\right),\ \left( 5, \ 9\right)\). Plot not shown. Shift right 2.
- Vertex \(\left( -1, \ 4\right)\); axis \(x=-1\); points \(\left( -4, \ -5\right),\ \left( -3, \ 0\right),\ \left( -2, \ 3\right),\ \left( 0, \ 3\right),\ \left( 1, \ 0\right),\ \left( 2, \ -5\right)\). Plot not shown. Reflect across the x-axis, shift left 1 and up 4.
5. Additional Learning Resources
These external resources support today's factoring and graphing skills. Use the indicated portions; later sections will extend factoring to expressions with non-unit leading coefficients.
| Resource and direct link | What it teaches | Scope / What to skip |
|---|---|---|
| OpenStax Elementary Algebra 2e §7.2 | Step-by-step factoring of \(x^2+bx+c\), including signs and worked checks; strong match for task 14–15 and Set 9–28. | Close match for the initial single-variable section. Skip the later \(x^2+bxy+cy^2\) two-variable section. |
| Khan Academy: Factoring simple quadratics review | A concise written explanation and examples of finding factors from their sums and products. | Close match. Focus on trinomials with coefficient 1 on \(x^2\). |
| Khan Academy: Factoring quadratics intro practice | Interactive practice rewriting trinomials and rectangle-area expressions as products. | Close match. Use for today's basic trinomial factoring. |
| OpenStax Intermediate Algebra 2e §6.3 | Illustrations of perfect-square trinomials and differences of squares, including conjugate products. | Some extra material. Read the perfect-square and difference-of-squares parts; skip sums and differences of cubes. |
These links use online resources; an internet connection is needed. Graphs in this file are embedded SVG, while displayed mathematical typesetting loads MathJax from a CDN.
6. Teacher / Parent Review — Source Packet
No critical mathematical error was found in the selected 2.7 task. The central sign patterns and factoring questions are consistent.
Modeling qualification — Main task problems 2–7 and 12–14. Subtracted side lengths \(x-k\) must be positive to represent physical rectangular sides. The packet does not explicitly state the relevant restrictions, so a student could mistakenly treat a negative numerical side length as a real rectangle. Suggested note: “Assume \(x\) is larger than every amount subtracted from it when using a physical area model; the algebraic identities themselves are valid for all real \(x\).”
Ready problem wording — numbers between roots/fractions. “Between” should be understood as strictly between; answers may use exact fractions or carefully justified decimals. This is an interpretation note, not a defect requiring correction.
7. Packet-to-Lesson Coverage Map
| Packet items | Skills required | Where taught here |
|---|---|---|
| Main task 1–7, 12 | Draw rectangles; expand \((x+m)(x+n)\) for positive/negative changes | In class, worked 1–2, signed area model |
| Main task 8, 13 | Determine signs of middle and constant terms | In class, worked 3 |
| Main task 9–11 | Opposite equal extensions; difference-of-squares identity | In class, worked 4–5 |
| Main task 14–15 | Factor monic quadratic trinomials | In class, worked 6–9 |
| Ready 1–8 | Three numbers between mixed numbers, negative numbers, fractions, radicals | Section 2, worked 10–15 |
| Set 9–28 | Factoring positive/negative trinomials, perfect squares, and differences of squares | In class, Section 3 |
| Go 29–32 | Graph parent, vertical shift, horizontal shift, reflection with three symmetric points on either side | Section 4, worked Go A–C and clipped plots |
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