Lesson 1.6 · Unit 1 · A-SSE.1a, A-SSE.1b, A-SSE.2
Seeing an expression as parts, and as a single object
There is a skill that separates students who find factoring easy from students who find it impossible, and it is not memorization. It is the ability to look at an expression and see its shape: to notice that something complicated has the form of something simple. That habit is worth building deliberately now, before you need it.
- Terms are separated by plus and minus signs at the top level. \( 3x^2 - 5x + 7 \) has three terms. The sign in front belongs to the term.
- Factors are separated by multiplication within a term. In \( 3x^2 \) the factors are 3, \( x \) and \( x \).
- The coefficient is the numerical factor of a term. In \( -5x \) the coefficient is \( -5 \), including the sign.
- A grouped quantity can be treated as a single object. In \( 4(x + 1)^2 - 9 \), the whole of \( (x + 1) \) behaves like one variable, so the expression has the shape "something squared minus a square".
- Look for a familiar form under the surface. \( a^2 - b^2 \), \( a^2 + 2ab + b^2 \), and a common factor in every term are the three shapes worth recognizing on sight.
- Rewriting reveals different information. The same quadratic in standard form shows the \( y \) intercept, in factored form shows the zeros, and in vertex form shows the maximum or minimum. None is more correct; they answer different questions.
- Ask what the question wants before choosing a form. Rewriting without a purpose is wasted work; rewriting toward the feature you were asked for is the whole strategy.
Where students lose marks: treating a coefficient as though it had no sign. In \( 7 - 4x \), the coefficient of \( x \) is \( -4 \), not 4. Stripping the sign off is what causes slope and direction errors later, when the sign is the whole answer.
The problem. Factor \( 16x^4 - 81 \) completely by recognizing its structure rather than by guessing.
Step one: count the terms and look for a common factor. There are two terms, \( 16x^4 \) and \( -81 \). They share no common factor other than 1, since 81 is odd and has no \( x \). So factoring out a GCF is not the route here.
Step two: ask whether each term is a perfect square. This is the question that unlocks the problem. \( 16x^4 = (4x^2)^2 \), because \( 4^2 = 16 \) and \( (x^2)^2 = x^4 \). And \( 81 = 9^2 \). Both terms are squares, and they are separated by a minus sign.
Step three: name the structure. The expression has the form \( a^2 - b^2 \) with \( a = 4x^2 \) and \( b = 9 \). Seeing this is the entire difficulty; the rest is mechanical.
Step four: apply the difference of squares pattern. Since \( a^2 - b^2 = (a - b)(a + b) \), \[ 16x^4 - 81 = (4x^2 - 9)(4x^2 + 9) \]
Step five: check whether either factor factors further. This is the step that the word "completely" is asking about, and it is the one students skip. Examine each factor in turn.
Step six: the first factor. \( 4x^2 - 9 \) is again a difference of squares, with \( 4x^2 = (2x)^2 \) and \( 9 = 3^2 \). So \( 4x^2 - 9 = (2x - 3)(2x + 3) \).
Step seven: the second factor. \( 4x^2 + 9 \) is a sum of squares, not a difference. A sum of squares does not factor over the real numbers, so this factor is left alone. Recognizing when to stop is as much a part of the skill as recognizing when to continue.
Step eight: write the complete factorization and verify. \[ 16x^4 - 81 = (2x - 3)(2x + 3)(4x^2 + 9) \] Check by multiplying the first two factors back: \( (2x - 3)(2x + 3) = 4x^2 - 9 \), then \( (4x^2 - 9)(4x^2 + 9) = 16x^4 - 81 \). The middle terms cancel in both products, which is exactly what the difference of squares pattern guarantees.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- How many terms does \( 5x^2 - 3x + 8 \) have?
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Three
- State the coefficient of \( x \) in \( 9 - 6x \).
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The sign belongs to the coefficient. \( -6 \)
- List the factors of the term \( 12x^3 \).
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12 and three factors of \( x \); the numerical factor 12 further factors as \( 2 \cdot 2 \cdot 3 \)
- Write \( x^2 - 25 \) in factored form.
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A difference of squares with \( a = x \), \( b = 5 \). \( (x - 5)(x + 5) \)
- Identify the common factor in \( 6x^2 + 9x \).
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\( 3x \)
- Factor \( 49y^2 - 16 \) and verify by expanding.
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\( 49y^2 = (7y)^2 \) and \( 16 = 4^2 \), so this is \( a^2 - b^2 \) with \( a = 7y \) and \( b = 4 \). It factors as \( (7y - 4)(7y + 4) \). Expanding: \( 49y^2 + 28y - 28y - 16 = 49y^2 - 16 \). The middle terms cancel, confirming it. \( (7y - 4)(7y + 4) \)
- The expression \( 3(x + 2)^2 - 12 \) has the structure of a difference of squares if you
first factor something out. Factor it completely.
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Both terms share a factor of 3: \( 3\big[(x + 2)^2 - 4\big] \). The bracket is now a difference of squares with \( a = (x + 2) \) and \( b = 2 \), so it factors as \( \big[(x + 2) - 2\big]\big[(x + 2) + 2\big] = (x)(x + 4) \). Altogether \( 3x(x + 4) \). Check by expanding: \( 3x^2 + 12x \), and the original expands to \( 3(x^2 + 4x + 4) - 12 = 3x^2 + 12x + 12 - 12 = 3x^2 + 12x \). They agree. \( 3x(x + 4) \)
- Explain what different information the forms \( x^2 - 6x + 8 \) and \( (x - 2)(x - 4) \)
each make immediately visible.
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They are the same expression written two ways, and each puts a different feature on the surface. Standard form shows the constant term 8, which is the value of the expression when \( x = 0 \), so it gives the \( y \) intercept of the graph at a glance. Factored form shows that the expression is zero exactly when \( x = 2 \) or \( x = 4 \), so it gives the \( x \) intercepts at a glance. Neither is simpler; you choose the one that answers the question being asked. Standard form shows the \( y \) intercept 8; factored form shows the zeros 2 and 4
- Explain why \( x^2 + 9 \) cannot be factored over the real numbers, while \( x^2 - 9 \)
can.
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The difference of squares pattern works because the middle terms cancel: \( (x - 3)(x + 3) = x^2 + 3x - 3x - 9 = x^2 - 9 \). For that cancellation to happen, the two factors must have opposite signs, and multiplying them then produces a minus sign in front of \( b^2 \). No pair of real factors gives \( +9 \) with no middle term, because any such pair would have to be \( (x + a)(x + b) \) with \( a + b = 0 \) and \( ab = 9 \), meaning \( b = -a \) and \( -a^2 = 9 \), which has no real solution. A sum of squares is therefore prime over the reals. A difference of squares needs factors of opposite sign, which produces a minus; no real pair gives \( +9 \) with no middle term
- Factor \( x^4 - 16 \) completely, and explain why stopping at two factors would be
incomplete.
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\( x^4 = (x^2)^2 \) and \( 16 = 4^2 \), so this is a difference of squares: \( x^4 - 16 = (x^2 - 4)(x^2 + 4) \). Stopping here would be incomplete because the first factor is itself a difference of squares: \( x^2 - 4 = (x - 2)(x + 2) \). The second factor, \( x^2 + 4 \), is a sum of squares and does not factor over the reals, so it stays. The complete factorization is \( (x - 2)(x + 2)(x^2 + 4) \). The word "completely" means every factor must be examined again after each step, and the process stops only when no factor can be broken down further. \( (x - 2)(x + 2)(x^2 + 4) \); the first factor of the two-factor form is itself a difference of squares