Special Factorizations: Chapter Review

Vocabulary and notation

Perfect square (of a term)

A number or term that is something squared: $49 = 7^2$, $9x^2 = (3x)^2$, $x^4 = (x^2)^2$.

Perfect-square trinomial

A trinomial that equals a squared binomial: one exactly when its first and last terms are perfect squares and its middle term is $\pm 2$ times the product of their roots.

Sum of squares $a^2 + b^2$

Two perfect squares joined by a plus. The difference-of-squares identity needs a minus, and a sum like $x^2 + 9$ has no real factorization at all, so a plus is a full stop for this identity.

SOAP

Same, Opposite, Always Positive: the mnemonic for the three signs of a cube factorization, read across the binomial's sign, the trinomial's middle sign, and the trinomial's last sign.

Conjugate

A two-term expression with the sign between its terms flipped: the conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$. Multiplying the two gives a difference of squares.

Rationalizing a denominator

Rewriting a fraction so no square root is left in the denominator, by multiplying by a form of $1$. The value never changes, only the way it is written.

Greatest common factor (GCF)

The largest factor every term shares, number and variable part together: the GCF of $3x^3$ and $12x$ is $3x$. Pull it out before testing any pattern.

Factoring completely

Factoring until no remaining factor can be factored again with integer coefficients.

Factoring by grouping

The method for a four-term polynomial: pair the terms, factor each pair, and remove the binomial both pairs leave behind.

Formulas and theorems

• Square of a sum

  $$(a + b)^2 = a^2 + 2ab + b^2$$

  Text description

  The square on a plus b splits into an a by a square, a b by b square, and two shaded a by b rectangles, and those two rectangles are the middle term 2ab.

  Use when All $a$ and $b$, either of which may be a term with a coefficient, in which case the whole term is squared.

  e.g. $(2x + 3)^2 = 4x^2 + 12x + 9$.

• Square of a difference

  $$(a - b)^2 = a^2 - 2ab + b^2$$

  Use when All $a$ and $b$. Only the middle term is negative; the last term is $+b^2$, since squaring $-b$ gives $+b^2$.

• Difference of squares

  $$a^2 - b^2 = (a + b)(a - b)$$

  Text description

  The square on a with a b by b corner removed cuts into two pieces that slide together into a rectangle of width a plus b and height a minus b.

  Use when All $a$ and $b$, but both terms must be perfect squares and the sign between them a minus. Which square is written first does not matter: $25 - 4x^2 = (5 + 2x)(5 - 2x)$.

  e.g. $9x^2 - 16 = (3x + 4)(3x - 4)$.

• Sum of cubes

  $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

  Use when All $a$ and $b$; unlike a sum of squares, this always factors. The trinomial carries a single $ab$, never $2ab$, and does not factor further over the integers.

  e.g. $8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9)$.

• Difference of cubes

  $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

  Use when All $a$ and $b$. The binomial's sign flips from the sum's, and the trinomial's middle sign flips with it; the last term stays $+b^2$.

  e.g. $x^3 - 64 = (x - 4)(x^2 + 4x + 16)$.

• Perfect cubes to recognize

  $$\begin{gathered} 1 = 1^3, \quad 8 = 2^3, \quad 27 = 3^3 \\ 64 = 4^3, \quad 125 = 5^3, \quad 216 = 6^3 \\ 1000 = 10^3 \end{gathered}$$

  Use when A monomial is a perfect cube when its coefficient is a perfect cube too: $8x^3 = (2x)^3$, $27x^3 = (3x)^3$. Cube-root the whole term, coefficient included, to read off $a$ and $b$.

• Clearing a single square root

  $$\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$$

  Use when $b > 0$. Multiply the numerator too.

• Conjugate product, one rational term

  $$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$

  Use when $b \ge 0$: the difference of squares with $\sqrt{b}$ as the second term. On a denominator it also needs $a^2 \neq b$, since $a^2 = b$ makes one of the two factors $0$.

  e.g. $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$.

• Conjugate product, two square roots

  $$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$$

  Use when $a \ge 0$ and $b \ge 0$; both squares collapse to plain numbers. On a denominator it also needs $a \neq b$.

  e.g. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) = 7 - 5 = 2$.

• A shared binomial factors out

  $$ka + kb = k(a + b)$$

  Use when Any expressions $k$, $a$, $b$; $k$ may be a whole binomial, not just a number.

• Splitting the middle term (the AC method)

  Two numbers with product $a \cdot c$ and sum $b$ split $bx$ into two terms, turning $ax^2 + bx + c$ into a four-term polynomial that groups.

  Use when $a \neq 0$, integer coefficients, and any overall common factor pulled out first. If no integer pair has both the right product and the right sum, the trinomial does not factor over the integers.

• Zero of a difference of squares

  $a^2 - b^2 = 0$ exactly when $a = b$ or $a = -b$: factor to $(a + b)(a - b) = 0$ and use the zero-product property.

  Use when Real $a$ and $b$. When $b \neq 0$ the two solutions are opposites, so this shape has two answers, not one.

  e.g. $x^2 - 9 = 0$ gives $(x + 3)(x - 3) = 0$, so $x = 3$ or $x = -3$.

• Mental-math shortcut

  A product of two numbers sitting the same distance $d$ either side of a round $n$ is $n^2 - d^2$.

  Use when Any $n$ and $d$. Worth spotting when a product is written as two numbers straddling a multiple of ten.

  e.g. $52 \times 48 = 50^2 - 2^2 = 2500 - 4 = 2496$.

Problem types, step by step

Exam traps

• Trap Squaring a sum term by term, $(a + b)^2 = a^2 + b^2$.

  Fix The cross-product $2ab$ is part of the answer and vanishes only when $a$ or $b$ is $0$: $(3 + 4)^2 = 49$ while $3^2 + 4^2 = 25$, and the missing $24$ is exactly $2ab$.

• Trap Squaring or rooting only the variable: $(2x)^2 = 2x^2$, or $\sqrt{9x^2} = 9x$.

  Fix The coefficient comes along: $(2x)^2 = 4x^2$ and $\sqrt{9x^2} = 3x$.

• Trap Giving both difference-of-squares factors the same sign: $a^2 - b^2 = (a - b)(a - b)$.

  Fix One plus and one minus. $(a - b)^2$ is the three-term $a^2 - 2ab + b^2$, a different expression.

• Trap Assuming a sum of cubes will not factor, by analogy with a sum of squares.

  Fix Only the sum of SQUARES refuses over the reals. A sum of cubes always factors: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.

• Trap Reading $a^3 + b^3$ as $(a + b)^3$.

  Fix They differ: $(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ has two extra middle terms.

• Trap Doubling the cube trinomial's middle term: $x^3 - 64 = (x - 4)(x^2 + 8x + 16)$.

  Fix The cube trinomial carries a single $ab$, so it is $(x - 4)(x^2 + 4x + 16)$. The doubled $2ab$ belongs to the perfect-square trinomial, a different pattern.

• Trap The sign trap in the second pair: factoring $+3$ out of $-3x + 6$ gives $3(2 - x)$, whose binomial does not match the $x - 2$ from the first pair.

  Fix Factor out $-3$ instead, so both pairs carry $x - 2$: $x^2(x - 2) - 3(x - 2) = (x - 2)(x^2 - 3)$.

• Trap Dropping the $1$ when a pair reduces to a bare binomial, reading $2x(x + 3) + (x + 3)$ as $(x + 3)(2x)$.

  Fix The bare binomial carries a coefficient of $1$: $2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)$.

• Trap Choosing the AC split by the sum alone, as with $10x = 5x + 5x$ for $3x^2 + 10x + 8$.

  Fix The pair must also multiply to $a \cdot c = 24$, and $5 \cdot 5 = 25$, so the working split is $6$ and $4$.

• Trap Stopping while a factor still factors, leaving $2(x^2 - 9)$ or $(x + 2)(4x^2 + 6)$ as the answer.

  Fix Finish both: $2(x + 3)(x - 3)$, and $2(x + 2)(2x^2 + 3)$.

• Trap Adding under a conjugate: $(\sqrt{6} - \sqrt{5})(\sqrt{6} + \sqrt{5}) = 6 + 5 = 11$.

  Fix The middle terms cancel and the squares subtract, giving $6 - 5 = 1$.