Factoring is multiplication run backward: instead of expanding a product, you rewrite a polynomial as a product of simpler pieces. This skill unlocks quadratic equations, because a product equals zero only when one of its factors does. In this chapter you will build a complete toolkit: GCF, grouping, trinomials, special patterns and the zero product property.
1. Greatest common factor (GCF)
The greatest common factor of the terms of a polynomial is the largest monomial that divides every term. To factor out the GCF, write it in front of parentheses and put the leftover quotients inside.
Always look for a GCF first, even when other methods will follow. Take the GCF of the coefficients (largest whole number dividing all of them), then each variable with its smallest exponent.
Factor \(18x^3 - 24x^2 + 12x\). The coefficients 18, 24 and 12 share the factor 6, and every term contains at least one \(x\). The GCF is \(6x\).
\[18x^3 - 24x^2 + 12x = 6x\,(3x^2 - 4x + 2)\]
Check by expanding: \(6x \cdot 3x^2 = 18x^3\), \(6x\cdot(-4x) = -24x^2\), \(6x \cdot 2 = 12x\).
If a term equals the GCF, a \(1\) must remain: \(5x^2 + 5x = 5x(x + 1)\), not \(5x(x)\). Always expand to check.
2. Factoring by grouping
When a polynomial has four terms, split it into two pairs, factor the GCF out of each pair, and hope that the same binomial shows up twice. Then factor that binomial out.
- Group the terms in two pairs.
- Factor the GCF out of each pair.
- If both pairs now contain the same binomial, factor it out as a common factor.
Factor \(x^3 + 4x^2 + 3x + 12\).
\[x^2(x + 4) + 3(x + 4) = (x + 4)(x^2 + 3)\]
Expanding \((x+4)(x^2+3)\) gives \(x^3 + 3x + 4x^2 + 12\), which matches the original polynomial.
If the signs of the second pair make the binomials opposite, factor out a negative GCF, for example \(-6x - 15\) becomes \(-3(2x + 5)\).
3. Trinomials \(x^2 + bx + c\)
A trinomial with leading coefficient 1 factors as \((x + p)(x + q)\) where the two numbers satisfy
\(x^2 + bx + c = (x + p)(x + q)\) exactly when \(p \cdot q = c\) and \(p + q = b\).
The area model above shows why: the rectangle of sides \(x+3\) and \(x+5\) is cut into pieces of area \(x^2\), \(5x\), \(3x\) and \(15\). The middle terms add up: \(5x + 3x = 8x\), so \((x+3)(x+5) = x^2 + 8x + 15\).
Factor \(x^2 - 2x - 35\). We need two numbers with product \(-35\) and sum \(-2\). The pair \(-7\) and \(5\) works.
\[x^2 - 2x - 35 = (x - 7)(x + 5)\]
When \(c\) is positive, \(p\) and \(q\) share the sign of \(b\). When \(c\) is negative, they have opposite signs and the one with the larger absolute value takes the sign of \(b\).
4. Trinomials \(ax^2 + bx + c\)
- Multiply \(a \cdot c\).
- Find two numbers with product \(ac\) and sum \(b\).
- Rewrite the middle term as the sum of two terms using those numbers.
- Factor by grouping.
Factor \(6x^2 + 11x + 3\). Here \(ac = 18\) and we need a sum of 11, so the numbers are 9 and 2.
\[6x^2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)\]
Factor \(3x^2 - 10x - 8\). Now \(ac = -24\) and the numbers with sum \(-10\) are \(-12\) and \(2\).
\[3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (x - 4)(3x + 2)\]
If no pair of integers works, the trinomial is called prime over the integers.
5. Special patterns
\(a^2 - b^2 = (a - b)(a + b)\). The picture shows the idea: remove a square of side \(b\) from a square of side \(a\) and the remaining area is \(a^2 - b^2\). A sum of squares such as \(a^2 + b^2\) does not factor over the integers.
\(a^2 + 2ab + b^2 = (a + b)^2\) and \(a^2 - 2ab + b^2 = (a - b)^2\).
\(49x^2 - 64 = (7x)^2 - 8^2 = (7x - 8)(7x + 8)\).
\(25x^2 - 30x + 9\): the first and last terms are \((5x)^2\) and \(3^2\), and the middle term is \(2 \cdot 5x \cdot 3 = 30x\). So \(25x^2 - 30x + 9 = (5x - 3)^2\).
On my planet we say: factor out the GCF first, then look for a pattern. After that, count the terms: 2 terms (difference of squares?), 3 terms (trinomial?), 4 terms (grouping?).
6. The zero product property
If \(A \cdot B = 0\), then \(A = 0\) or \(B = 0\) (or both). This works for any number of factors, but only when the product equals zero.
The parabola above is \(y = (x-2)(x-3)\). It crosses the x-axis where a factor is zero, at \(x = 2\) and \(x = 3\). Factoring therefore tells you where a graph meets the x-axis.
7. Solving equations by factoring
- Move every term to one side so the other side is 0.
- Factor the polynomial completely.
- Set each factor equal to 0 and solve.
- Check each solution in the original equation.
Solve \(x^2 + 6x = 16\).
\[x^2 + 6x - 16 = 0 \;\Rightarrow\; (x + 8)(x - 2) = 0 \;\Rightarrow\; x = -8 \text{ or } x = 2\]
Check: \((-8)^2 + 6(-8) = 64 - 48 = 16\) and \(2^2 + 6 \cdot 2 = 16\).
Never divide both sides by \(x\) to simplify \(x^2 = 7x\): you lose the solution \(x = 0\). Write \(x^2 - 7x = 0\), then \(x(x - 7) = 0\). Also, \((x - 2)(x + 3) = 14\) does not mean \(x - 2 = 14\); the property only applies to a product of 0.
Key takeaways
- Always factor out the greatest common factor first, and check by expanding.
- Four terms: try grouping. Three terms: find two numbers with product \(c\) (or \(ac\)) and sum \(b\).
- \(a^2 - b^2 = (a-b)(a+b)\); \(a^2 \pm 2ab + b^2 = (a \pm b)^2\).
- Zero product property: \(AB = 0\) means \(A = 0\) or \(B = 0\).
- To solve by factoring, set one side to 0, factor, set each factor to 0, then check.
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