Last updated .
2.8H — The Wow Factor
Read the explanations in order, work through the examples with pencil and paper, then complete the corresponding section of the packet. This lesson uses the factoring ideas from 2.6–2.7; it does not require lessons 2.9 or 2.10.
1. In class: A Solidify Understanding Task
From rectangle dimensions to quadratic expressions
Recall that the area of a rectangle is length × width. A rectangle with side lengths \(x+1\) and \(x+3\) has area \((x+1)(x+3)\). Multiplying gives \(x^2+4x+3\). Factoring reverses that multiplication.
Suppose a customer orders two copies of the rectangle with dimensions \(x+1\) and \(x+3\). Its total area is
\[2(x+1)(x+3)=2(x^2+4x+3)=2x^2+8x+6.\]The leading coefficient is 2 because there are two identical rectangles. Keep the 2 outside the parentheses when you want the number of copies to be visible.
A different situation gives the same sort of expression: a single rectangle may have dimensions \(2x+2\) and \(x+3\). Then \((2x+2)(x+3)=2(x+1)(x+3)\). The picture differs even though the total area formula is equivalent.
A single rectangle with a multiple of x in one side
Instead of two copies, suppose a rectangular tile layout has sides \(2x+1\) and \(x+4\):
\[(2x+1)(x+4)=2x^2+8x+x+4=2x^2+9x+4.\]The leading term \(2x^2\) tells us that at least one side can involve a multiple of x. The coefficient of the middle term comes from the two cross-products.
Expand \((3x+2)(2x+5)\).
\[(3x)(2x)+(3x)(5)+2(2x)+2(5)\] \[=6x^2+15x+4x+10=6x^2+19x+10.\]Notice how the leading coefficient is \(3\cdot2=6\), the constant is \(2\cdot5=10\), and the middle coefficient is \(15+4=19\).
Factoring when the coefficient of x² is not 1
The two-number method from 2.7 worked directly on \(x^2+bx+c\). For \(ax^2+bx+c\), where \(a\) is not 1, we must account for the leading coefficient.
Method 1 — Factor a greatest common factor first
A greatest common factor (GCF) is a factor shared by every term.
Now find two numbers with product 12 and sum 7: 3 and 4.
\[3(x^2+7x+12)=\boxed{3(x+3)(x+4)}.\]Check: \((x+3)(x+4)=x^2+7x+12\); multiplying by 3 returns the original expression.
The middle term is zero because the cross-products cancel: \(3x-3x=0\).
Method 2 — The ac method (split and group)
To factor \(ax^2+bx+c\) with integer coefficients, look for two integers \(m,n\) such that
\[mn=ac\qquad\text{and}\qquad m+n=b.\]Replace \(bx\) with \(mx+nx\), then factor the four terms in pairs. This works because we are creating two groups with a shared binomial factor.
Factor \(6x^2+7x+2\).
\[ac=6\cdot2=12;\qquad 3+4=7,\quad3\cdot4=12.\] \[6x^2+7x+2=6x^2+3x+4x+2\] \[=3x(2x+1)+2(2x+1)\] \[=\boxed{(3x+2)(2x+1)}.\]Check: \(6x^2+3x+4x+2=6x^2+7x+2\).
Factor \(4x^2-8x+3\).
\[ac=12;\qquad(-6)+(-2)=-8,\quad(-6)(-2)=12.\] \[4x^2-8x+3=4x^2-6x-2x+3\] \[=2x(2x-3)-1(2x-3)\] \[=\boxed{(2x-1)(2x-3)}.\]Both constants are negative because the constant term is positive and the middle coefficient is negative.
Factor \(3x^2+x-10\).
\[ac=-30;\qquad 6+(-5)=1,\quad 6(-5)=-30.\] \[3x^2+x-10=3x^2+6x-5x-10\] \[=3x(x+2)-5(x+2)=\boxed{(3x-5)(x+2)}.\]Because \(c\) is negative, the constants in the linear factors must have opposite signs (when the leading coefficient is positive and the factors are written with positive leading terms).
A known side helps you find the other side
The area is \(5x^2+8x+3\), and one side is \(x+1\). Which other side makes the area?
Because the first term must be \(5x^2\), try \(5x\) in the other factor. The constant terms must multiply to 3; with \(+1\) already present, try \(+3\).
\[(x+1)(5x+3)=5x^2+3x+5x+3=5x^2+8x+3.\]Therefore the other side is \(\boxed{5x+3}\).
Factor \(10x^2+17x+3\). Find \(ac=30\), with \(15+2=17\).
\[10x^2+17x+3=10x^2+15x+2x+3\] \[=5x(2x+3)+1(2x+3)=\boxed{(5x+1)(2x+3)}.\]Check: \(10x^2+15x+2x+3\) agrees with the original.
Reading the layout, not just the expression
In a rectangle model, \(x^2\) tiles represent squares of side x, the \(x\)-tiles represent rectangles of area x, and unit squares represent area 1. To construct a single rectangle, place them so the entire outer boundary is rectangular and read its side lengths. Different arrangements of the same collection of tiles can produce different outer dimensions or may leave gaps; verify any proposed dimensions algebraically.
2. Ready review — Arithmetic versus geometric sequences
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 packet gives the first and fifth terms and asks for the missing three terms. Four equal steps connect term 1 to term 5.
| Sequence | What stays constant? | First-to-fifth relationship |
|---|---|---|
| Arithmetic | Add a common difference \(d\) | \(a_5=a_1+4d\), so \(d=(a_5-a_1)/4\) |
| Geometric | Multiply by a common ratio \(r\) | \(a_5=a_1r^4\), so \(r^4=a_5/a_1\) (if \(a_1\ne0\)) |
The first term is 6 and the fifth term is 30.
\[d=\frac{30-6}{4}=6.\]Terms 1–5: \(\boxed{6,12,18,24,30}\). Four additions of 6 produce the fifth term.
The first term is 2 and the fifth term is 162.
\[r^4=\frac{162}{2}=81.\]Taking a positive common ratio gives \(r=3\): \(\boxed{2,6,18,54,162}\). But \(r=-3\) also works: \(\boxed{2,-6,18,-54,162}\). Both have the same first and fifth terms.
The first and fifth terms alone do not determine a unique real geometric sequence when the fifth-to-first ratio is positive. Both positive and negative ratios can give the same fifth term because \((-r)^4=r^4\). If the intended task is to find a unique geometric sequence, it must specify a positive common ratio (or some equivalent restriction). When no restriction is stated, show both possibilities.
Check your work — Ready 1–3 answer key
- Missing arithmetic terms: \(471,\ 939,\ 1407\) (difference \(468\)). Missing geometric terms: \(15,\ 75,\ 375\) (ratio \(5\)), or \(-15,\ 75,\ -375\) (ratio \(-5\)). Both ratios fit the given first and fifth terms.
- Missing arithmetic terms: \(-1098,\ -738,\ -378\) (difference \(360\)). Missing geometric terms: \(-486,\ -162,\ -54\) (ratio \(\frac{1}{3}\)), or \(486,\ -162,\ 54\) (ratio \(- \frac{1}{3}\)). Both ratios fit the given first and fifth terms.
- Missing arithmetic terms: \(769,\ 514,\ 259\) (difference \(-255\)). Missing geometric terms: \(256,\ 64,\ 16\) (ratio \(\frac{1}{4}\)), or \(-256,\ 64,\ -16\) (ratio \(- \frac{1}{4}\)). Both ratios fit the given first and fifth terms.
3. Set — Area models and additional factoring practice
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.
For the area diagrams (problems 4–7), count the \(x^2\) squares, the \(x\)-rectangles, and the unit tiles to write the expanded expression. Then read the entire rectangle's length and width to write the factored form. Compare the arrangements rather than assuming that identical tile counts mean identical dimensions.
A rectangle has dimensions \(2x+3\) and \(2x+1\). Write its area both ways.
\[A=(2x+3)(2x+1)\] \[=4x^2+2x+6x+3=\boxed{4x^2+8x+3}.\]The four area contributions are \(4x^2\), \(2x\), \(6x\), and 3.
Factor \(4x^2+7x-2\). Here \(ac=-8\); choose 8 and −1, which sum to 7.
\[4x^2+8x-x-2=4x(x+2)-1(x+2)\] \[=\boxed{(4x-1)(x+2)}.\]To check your Set factoring, expand each answer and compare the three coefficients with the original. When an expression has two terms, check for a GCF or difference of squares before trying a four-term grouping method.
Check your work — Set 4–23 answer key
- \((3x+2)(2x+3)=6x^2+13x+6\). Using the unit strips and unit squares indicated by the tile convention.
- \((2x+2)(3x+3)=6x^2+12x+6\). Using the unit strips and unit squares indicated by the tile convention.
- \((6x+3)(x+2)=6x^2+15x+6\). Using the unit strips and unit squares indicated by the tile convention.
- (A) The rectangles have different side lengths and different numbers of \(x\) strips (13, 12, and 15). (B) The \(x^2\) and constant coefficients remain 6; the middle coefficients differ. (C) The side lengths determine the factors: \((3x+2)(2x+3)\), \((2x+2)(3x+3)\), and \((6x+3)(x+2)\).
- \(\left(x + 2\right) \left(4 x - 1\right)\).
- \(\left(x - 5\right) \left(2 x + 3\right)\).
- \(\left(2 x + 3\right) \left(3 x - 1\right)\).
- \(\left(x - 1\right) \left(4 x + 3\right)\).
- \(\left(x + 5\right) \left(4 x - 1\right)\).
- \(\left(x - 2\right) \left(3 x - 4\right)\).
- \(\left(2 x - 1\right) \left(3 x + 2\right)\).
- \(\left(x - 6\right) \left(3 x + 4\right)\).
- \(\left(x + 2\right) \left(2 x + 5\right)\).
- \(\left(x + 6\right) \left(5 x + 1\right)\).
- \(\left(x + 2\right) \left(5 x - 3\right)\).
- \(\left(2 x - 1\right) \left(2 x + 5\right)\).
- \(3 \left(x - 5\right) \left(x + 5\right)\).
- \(\left(x + 2\right) \left(3 x + 1\right)\).
- \(\left(2 x - 1\right) \left(2 x + 5\right)\).
- \(\left(x + 2\right) \left(2 x - 3\right)\).
4. Go review — Finding a parabola’s axis of symmetry from its x-intercepts
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 x-intercepts are the places a graph crosses the x-axis, and a parabola's axis of symmetry passes halfway between two distinct x-intercepts. If they are at \((r_1,0)\) and \((r_2,0)\), average their x-coordinates:
This is the midpoint of the two x-intercepts along the x-axis. The answer is an equation for a vertical line, not just a number.
A parabola has x-intercepts at \((-5,0)\) and \((7,0)\). Find the axis of symmetry.
\[x=\frac{-5+7}{2}=\frac{2}{2}=\boxed{1}.\]The required equation is \(\boxed{x=1}\). It is 6 units from each intercept.
The x-intercepts are \((-\tfrac{7}{4},0)\) and \((\tfrac{5}{2},0)\). Find the axis of symmetry.
\[x=\frac{-\frac74+\frac52}{2}=\frac{-\frac74+\frac{10}4}{2}=\frac{\frac34}{2}=\boxed{\frac38}.\]So the line of symmetry is \(\boxed{x=\frac38}\). The common mistake is forgetting the final division by 2.
Find the axis when the intercepts are \((-11,0)\) and \((-3,0)\).
\[x=\frac{-11+(-3)}2=\frac{-14}2=-7.\]The line is \(\boxed{x=-7}\). Being between two negative numbers does not make the midpoint positive.
Check your work — Go 24–29 answer key
- \(x=0\).
- \(x=6\).
- \(x=\frac{3}{2}\).
- \(x=- \frac{17}{2}\).
- \(x=25\).
- \(x=\frac{3}{4}\).
Additional Learning Resources
These pages supplement the textbook-style explanations above. Use only the specified topics so you do not jump ahead unnecessarily.
| Resource and link | What it teaches | Scope / what to skip |
|---|---|---|
| Khan Academy — Factoring quadratics with leading coefficient ≠ 1 | Clear walkthrough of middle-term splitting and grouping for \(ax^2+bx+c\). | Close match. Read the full article and try its practice checks. |
| OpenStax — 6.2 Factor Trinomials | Expanded textbook explanation and worked examples, especially “Factor Trinomials … using the ac Method.” | Some extra material. Focus on the ac-method subsection; skip the later substitution problems. |
| OpenStax — 9.2 Arithmetic Sequences | How to find missing terms using a common difference, including known first and later terms. | Some extra material. Use the opening examples and finding terms from a known term; skip applications not needed for Ready. |
| OpenStax — 12.3 Geometric Sequences and Series | Common ratios and geometric terms. | Significantly broader. Read the sequence definitions and early examples only; skip geometric series and sums. |
The links above are external resources. Mathematical notation on this page uses MathJax loaded from a CDN, so an internet connection is needed for formatted equations.
Teacher / Parent Review — Material packet issues
2.8H Ready, problems 1–3: As written, the first and fifth terms permit both positive and negative real common ratios whenever their ratio is positive. Clarify “use a positive common ratio” if one answer per row is intended; otherwise accept both.
2.8H main task, page 47: The final instruction “Write a recipe” repeats problem number 12. Renumber it 13 for unambiguous reference.
2.8H main task, questions involving quilt “multiples”: Algebraically equivalent expressions can describe multiple identical blocks or one rectangle with scaled dimensions. Consider specifying whether a physical arrangement or only the area formula is requested.
Packet-to-lesson coverage
| Original section | Skills assessed | Covered here |
|---|---|---|
| Main task 1–5 | Area models, external multipliers, GCF, expanding/factoring | In class, worked examples 1–4 |
| Main task 6–10 | Finding missing side, factoring when leading coefficient is non-unit, signs | In class, worked examples 5–8 |
| Main task 11–final “recipe” | Both side lengths with multiples of x; general ac method | In class, worked examples 2 and 9 |
| Ready 1–3 | Arithmetic vs. geometric sequences, first/fifth terms | Section 2, worked examples 10–11 and clarification |
| Set 4–7 | Tile counts, area models, equivalent expressions | Section 3, worked example 12 and diagram |
| Set 8–23 | General trinomial factoring, GCF and differences of squares | In class and Section 3, worked examples 3–9 and 13 |
| Go 24–29 | Axis of symmetry from two x-intercepts, including fractions and negatives | Section 4, Go problems A–C and number line |
Curriculum: Mathematics Vision Project, Secondary Math II, Module 2, 2017, Section 2.8H, CC BY 4.0. This supplementary explanation is independently written to accompany that section.
Last updated .