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2.6 — Factor Fixin’
Earlier lessons used squares to understand quadratic expressions. A rectangle whose sides are different lengths gives another important pattern. For today's new factoring method, we begin with rectangles whose sides are \(x+m\) and \(x+n\), where the added lengths are integers. The order is deliberate: build the area first, then learn to recover the side lengths from the area.
1. In class: A Develop Understanding Task
A rectangular area has two equivalent expressions
Start with an \(x\)-by-\(x\) square. Extend one dimension by 4 units and the other by 2 units. Its new dimensions are \(x+4\) and \(x+2\), so its area is \((x+4)(x+2)\).
Multiply each part of \((x+4)(x+2)\):
\[\begin{aligned}(x+4)(x+2)&=x\cdot x+x\cdot2+4\cdot x+4\cdot2\\&=x^2+2x+4x+8\\&=x^2+6x+8.\end{aligned}\]Thus \(A(x)=(x+4)(x+2)=x^2+6x+8\). The product tells us the side lengths; the expanded expression shows the extra area beyond the original \(x^2\): \(6x+8\).
There are two strips with combined area \((m+n)x\) and a corner with area \(mn\). This identity works for any real \(m,n\); positive extensions are the natural context for today's rectangle models.
If one side is unchanged and the other gains 7, then \(A=x(x+7)=x^2+7x\): there is no constant corner area. If both sides gain 5, then \(A=(x+5)^2=x^2+10x+25\); this is the perfect-square pattern from Section 2.3.
Reverse multiplication: find the side lengths
Factoring means expressing a polynomial as a product. Suppose a rectangle has area \(x^2+9x+20\). To find the side additions, look for two numbers that have sum 9 and product 20. The pair is 4 and 5.
Check: \(4+5=9\) creates the middle coefficient, while \(4\cdot5=20\) creates the constant term. The rectangle's sides are \(x+4\) and \(x+5\).
A systematic search rather than guessing
List the factor pairs of \(c\), then check their sums. For \(x^2+13x+36\), the positive factor pairs of 36 are \((1,36),(2,18),(3,12),(4,9),(6,6)\). Their sums are 37, 20, 15, 13, and 12. The successful pair is 4 and 9.
Factor \(x^2-7x-30\). Since the constant is negative, one factor must be positive and the other negative. The numbers \(-10\) and \(3\) multiply to \(-30\) and sum to \(-7\).
\[x^2-7x-30=(x-10)(x+3).\]Check: \(x^2+3x-10x-30=x^2-7x-30\). Algebraically this is valid for all real \(x\), although the geometric side length \(x-10\) is positive only for \(x>10\).
Factor \(x^2-11x+24\). Two negative numbers must multiply to positive 24 and add to negative 11: \(-3\) and \(-8\).
\[x^2-11x+24=(x-3)(x-8).\]Why identical outside coefficients can give different middle coefficients
Notice that \(x^2+10x+24=(x+4)(x+6)\), while \(x^2+11x+24=(x+3)(x+8)\) and \(x^2+14x+24=(x+2)(x+12)\). Each pair has the same product, 24, but a different sum. That is why the \(x^2\) coefficient and constant term can stay fixed while the \(x\) coefficient changes.
The area \(x^2+7x+9\) cannot be factored as \((x+m)(x+n)\) with integer extensions, because no two integers add to 7 and multiply to 9. In that intended integer-extension model, the shop cannot use two whole-number additions.
This does not show that no rectangle is possible: \(x^2+7x+9\) is positive for \(x\ge0\), so one can construct a rectangle of area \(x^2+7x+9\), for example with side lengths \(1\) and \(x^2+7x+9\). Moreover, the expression has real linear factors \(\bigl(x+\frac{7+\sqrt{13}}2\bigr)\bigl(x+\frac{7-\sqrt{13}}2\bigr)\). The packet's intended restriction to integer extensions needs to be stated.
2. Ready review — Common factors and difference of squares
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 Ready exercises introduce two additional ways of expressing quadratics as products. Both start with familiar multiplication rules.
Factor out a shared monomial
Factor \(15x^2-25x\). Both terms contain \(5x\), so take it out:
\[15x^2-25x=5x(3x-5).\]Check by distributing: \(5x\cdot3x=15x^2\) and \(5x\cdot(-5)=-25x\). There is no constant term in this quadratic: \(c=0\).
Multiply conjugates to get a difference of squares
The pair \((u+v)\) and \((u-v)\) are called conjugates. Their cross-products cancel.
Notice \(b=0\). More generally, \((u+v)(u-v)=u^2-v^2\), called a difference of squares.
Check your work — Ready 1–13 answer key
- \(4x^2-7x\).
- \(15x^2+40x\).
- \(9x^2-6x\).
- \(c=0\).
- \(x(x+4)\).
- \(7x(x-3)\).
- \(12x(x+5)\).
- \(4x(2x+5)\).
- \(x^2-81\).
- \(x^2-4\).
- \(36x^2-25\).
- \(49x^2-1\).
- \(b=0\).
3. Set review — Factoring organized by a fixed constant
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 homework problems 14–25, the leading coefficient is 1. Organize your search by factor pairs. Even if the last term matches in three successive problems, you must test the required sum separately each time.
Factor each expression.
\[\begin{aligned}x^2+9x+18&=(x+3)(x+6),\\x^2+11x+18&=(x+2)(x+9),\\x^2+19x+18&=(x+1)(x+18).\end{aligned}\]All three products have constant 18; the different pairs of side extensions make the middle coefficients 9, 11, and 19.
Factor \(x^2+16x+64\). Since \(8+8=16\) and \(8\cdot8=64\), the factors are equal:
\[x^2+16x+64=(x+8)^2.\]Not every trinomial with a square constant is itself a perfect-square trinomial. The middle coefficient must be exactly twice the square root of the constant (with the appropriate sign).
Check your work — Set 14–26 answer key
- \(\left(x + 5\right) \left(x + 9\right)\).
- \(\left(x + 3\right) \left(x + 15\right)\).
- \(\left(x + 1\right) \left(x + 45\right)\).
- \(\left(x + 3\right) \left(x + 8\right)\).
- \(\left(x + 4\right) \left(x + 6\right)\).
- \(\left(x + 2\right) \left(x + 12\right)\).
- \(\left(x + 6\right)^{2}\).
- \(\left(x + 4\right) \left(x + 9\right)\).
- \(\left(x + 2\right) \left(x + 18\right)\).
- \(\left(x - 20\right) \left(x + 5\right)\).
- \(\left(x + 10\right)^{2}\).
- \(\left(x + 4\right) \left(x + 25\right)\).
- For \((x+p)(x+q)=x^2+(p+q)x+pq\), different factor pairs can share product \(pq=c\) but have different sums \(p+q=b\). In the last row, #23 actually has \(c=-100\), whereas #24–25 have \(c=100\); the packet’s claim that every row has the same outside coefficients is false there.
4. Go review — Square roots of perfect squares
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 principal square root \(\sqrt{N}\) means the nonnegative number whose square is \(N\). Consequently \(\sqrt{U^2}=|U|\), not always \(U\). Absolute-value bars are necessary whenever \(U\) could be negative.
Evaluate \(\sqrt{(5x+2)^2}\).
\[\sqrt{(5x+2)^2}=|5x+2|.\]For instance, if \(x=-1\), the inside expression is \((-3)^2=9\), whose principal square root is 3; \(|5(-1)+2|=3\), while \(5(-1)+2=-3\) is incorrect.
Simplify \(\sqrt{x^2-18x+81}\).
Half of \(-18\) is \(-9\), and \((-9)^2=81\), so the radicand is \((x-9)^2\).
\[\sqrt{x^2-18x+81}=\sqrt{(x-9)^2}=|x-9|.\]Check: the square of \(|x-9|\) returns \(x^2-18x+81\).
Simplify \(\sqrt{144m^2n^6}\).
\[144m^2n^6=(12mn^3)^2\quad\Longrightarrow\quad\sqrt{144m^2n^6}=|12mn^3|=12|m|\,|n|^3.\]Do not silently assume variables are nonnegative. If the homework or teacher explicitly assumes positive variable values, the absolute values may then be omitted.
Decide whether \(\sqrt{x^2+12x+32}\) is the square root of a perfect-square trinomial.
A square with middle term \(12x\) would require constant \((12/2)^2=36\), but the constant here is 32. In the packet's instructions, leave the entire expression under the radical sign: \(\sqrt{x^2+12x+32}\). Do not force a simplification.
The instruction to “take the square root” requires the principal square root. In particular, problem 28 gives \(\sqrt{(3x-7)^2}=|3x-7|\), and the complete-square trinomials in problems 30, 31, 34, and 35 also need absolute-value signs unless a sign assumption about \(x\) is supplied. Problem 27 has an entire squared expression; problem 29 has \(121a^2b^6\), which equals \((11ab^3)^2\). For the other radicands, follow the packet's instruction to leave non-perfect-square expressions unsimplified.
Check your work — Go 27–35 answer key
- \(17|xyz|\). Perfect square; this is its principal square root.
- \(|3x-7|\). Perfect square; this is its principal square root.
- \(11|ab^3|\). Perfect square; this is its principal square root.
- \(|x+4|\). Perfect square; this is its principal square root.
- \(|x+7|\). Perfect square; this is its principal square root.
- \(\sqrt{x^2+14x-49}\). Not a perfect-square polynomial; leave under the radical as requested.
- \(\sqrt{x^2+10x+100}\). Not a perfect-square polynomial; leave under the radical as requested.
- \(|x+10|\). Perfect square; this is its principal square root.
- \(|x-10|\). Perfect square; this is its principal square root.
Additional Learning Resources
These are optional supplements to this lesson, not prerequisites for later sections. The links below were checked for their relevant content.
| Resource and direct link | What it teaches | Scope / What to skip |
|---|---|---|
| Khan Academy — Factoring quadratics, leading coefficient 1 | Clear sum-and-product method with examples; matches the main task and Set. | Close match. Focus on factors of \(x^2+bx+c\). |
| OpenStax — 7.2 Factor Trinomials of the Form \(x^2+bx+c\) | Written explanations and worked practice factoring monic trinomials. | Some extra material. Read the single-variable trinomial section; skip the later two-variable forms \(x^2+bxy+cy^2\). |
| Khan Academy — Factoring quadratics intro (practice) | Interactive factoring with rectangular area contexts. | Close match. Practice the introductory exercises; no need to jump to solving equations by factoring. |
Teacher / parent review — Material issues only
Main task, problem 18. The question asks whether an area \(x^2+7x+9\) can form a rectangle; the intended answer appears to require extensions by whole-number amounts but that assumption is unstated. Without it the blanket claim “cannot make the rectangle” is false. Clarify: “Can it be made by adding nonnegative integer amounts to the two sides of the original \(x\)-by-\(x\) square?”
Go, problems 27–35. A correct principal-square-root answer sometimes requires absolute values. Model \(\sqrt{u^2}=|u|\), or explicitly specify nonnegative expressions if that is the convention desired. The latter should not be taken for granted.
Section 2.6 packet-to-lesson coverage
| Packet task | Skills / problem numbers | Where taught |
|---|---|---|
| Main task | Rectangular area, expanded and factored forms, diagrams (1–9) | In class and area diagram |
| Main task | Reverse engineering integer side extensions, sum-product rule, limitations (10–18) | In class and clarification |
| Ready | Distribute monomial × binomial; identify \(c=0\); take common factor (1–8) | Section 2 |
| Ready | Conjugates / difference of squares; identify \(b=0\) (9–13) | Section 2 |
| Set | Factor monic trinomials, positive and negative constants, fixed \(c\) and varying \(b\) (14–26) | In class and Section 3 |
| Go | Identify perfect-square radicands, principal roots, and when to leave radicals unchanged (27–35) | Section 4 and Go A–D |
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