5.2 Solving Quadratics by Factoring

5.2 Solving Quadratics by Factoring

Table 11 lists solving via factoring or the quadratic formula as a single bullet, but AAF items often tell you the method by the structure of the options: integer roots suggest factoring; messy radicals suggest the formula. This section is factoring as a solving tool. The algebraic factoring techniques themselves — GCF, grouping, special products, and ac for trinomials — live in Factoring Quadratic Trinomials, with GCF and special products at GCF, Grouping, and Special Products.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Zero product property

If A · B = 0, then A = 0 or B = 0 (or both). That is the entire reason factoring solves equations. It does not apply to A · B = 12. Never split x(x + 3) = 12 into x = 12 or x + 3 = 12. Those two guesses fail: 12 · 15 = 180, not 12.

The workflow is:

1. Write in standard form. Move every term to one side so the equation is ax^2 + bx + c = 0.
2. Factor (GCF first, then the remaining quadratic).
3. Set each factor equal to zero.
4. Check in the original equation if the problem later involves a denominator or a square root; for a plain quadratic polynomial, both real roots should work.

Standard form is not optional. x^2 + 5x = 14 is not factored until 14 has moved. 2x^2 = 8x is not solved by dividing by 2x.

Worked example: x^2 + 5x = 14 → x^2 + 5x − 14 = 0

The given equation is not yet standard form. Subtract 14:

x^2 + 5x − 14 = 0

Find two integers whose product is −14 and whose sum is 5: 7 and −2, because 7 · (−2) = −14 and 7 + (−2) = 5.

(x + 7)(x − 2) = 0

Zero product:

x + 7 = 0 ⇒ x = −7

x − 2 = 0 ⇒ x = 2

Two distinct real roots: x = −7 and x = 2.

Check: (−7)^2 + 5(−7) = 49 − 35 = 14. 2^2 + 5(2) = 4 + 10 = 14. Both recover the original right-hand side.

If you had factored as (x + 2)(x − 7), the sum of 2 and −7 is −5, not +5. Sign errors are the main CAT trap on this item type. The pair 14 and −1 multiplies to −14 but sums to 13, not 5 — do not grab factors of 14 without checking the sum against b.

The same equation as a graph

y = x^2 + 5x − 14 is a parabola opening up. The solutions of y = 0 are the x-intercepts −7 and 2. The axis is x = (−7 + 2)/2 = −2.5, matching x = −b/(2a) = −5/2. Solving by factoring is therefore the same skill as finding intercepts from intercept form. Skills Insight 250–262 still wants that graph connection even when the stem never draws the parabola.

Two real roots, a double root, and no real factorization over the reals

Double root in more detail. Graphically the vertex sits on the x-axis. AAF may list the solution as x = 3 or as a repeated root; do not invent a second distinct number such as x = −3. Expanding (x − 3)^2 is x^2 − 6x + 9, not x^2 − 9.

No real factorization. There is no real x. Switch to the quadratic formula in the next section if you need to confirm the discriminant; on AAF, the intended choice is often no real solution. Complex roots are not the emphasis of this placement test.

Not every real-root quadratic factors nicely over the integers. x^2 − 2 = 0 is (x − √2)(x + √2) = 0. If integer factoring stalls after a short attempt, use the formula rather than forcing fake integer pairs such as (x − 2)(x + 1).

Leading coefficient not 1

Solve 2x^2 + 7x − 15 = 0. Numbers that multiply to 2 · (−15) = −30 and add to 7 are 10 and −3.

2x^2 + 10x − 3x − 15 = 0

2x(x + 5) − 3(x + 5) = 0

(2x − 3)(x + 5) = 0

2x − 3 = 0 ⇒ x = 3/2

x + 5 = 0 ⇒ x = −5

Both are real. Leave 3/2 as a fraction. A calculator decimal 1.5 may appear among the options, but exact form is safer unless every option is decimal.

GCF first: 4x^2 − 8x = 0 is 4x(x − 2) = 0, so x = 0 or x = 2. If you cancel 4x from 4x^2 = 8x by dividing both sides by 4x, you lose x = 0. That is the signature trap of this chapter.

Trap: dividing both sides by x and losing x = 0

Wrong path: x^2 = 5x. Divide both sides by x to get x = 5. That discards the solution x = 0, which does satisfy 0 = 0.

Correct path: x^2 − 5x = 0 → x(x − 5) = 0 → x = 0 or x = 5.

The same error appears as x^2 + 3x = 0 divided by x to leave x + 3 = 0. Always move to = 0 and factor out x instead of dividing by the variable. On a CAT, the option that lists only the nonzero root is written to look complete.

A related trap: dividing 2x^2 = 8 by 2 is legal (x^2 = 4) because 2 is a nonzero constant. Then x = ±2, not x = 2 only. Dividing by an expression that might be zero is the problem; dividing by a nonzero constant is ordinary simplification.

Another fully worked integer case

Solve 3x^2 − 14x = 5.

Standard form: 3x^2 − 14x − 5 = 0.

Product 3 · (−5) = −15, sum −14: the pair is −15 and 1.

3x^2 − 15x + x − 5 = 0

3x(x − 5) + 1(x − 5) = 0

(3x + 1)(x − 5) = 0

x = −1/3 or x = 5.

Check: 3(25) − 14(5) = 75 − 70 = 5. 3(1/9) − 14(−1/3) = 1/3 + 14/3 = 15/3 = 5. Both work. If the stem asked only for a positive length, you would keep x = 5; if it asked for solutions of the equation, you keep both.

Incomplete quadratics

No linear term: x^2 − 16 = 0 is a difference of squares: (x − 4)(x + 4) = 0, so x = ±4. Equivalently x^2 = 16 and take both square roots. Forgetting the negative root is as costly as dropping x = 0. Difference of squares is reviewed with special products in the factoring chapter.

No constant term: x^2 + 8x = 0 → x(x + 8) = 0 → x = 0 or x = −8. This is the family that tempts division by x.

Perfect square: (2x − 1)^2 = 0 has the single root x = 1/2. Expanding first is optional: 4x^2 − 4x + 1 = 0 factors back to the same square. Setting 2x − 1 = ±0 still yields one value.

Creating, then factoring

A number is 6 more than another, and their product is 40. Let the smaller be n:

n(n + 6) = 40

n^2 + 6n − 40 = 0

(n + 10)(n − 4) = 0

n = −10 or n = 4. In a length context you discard −10 if the problem requires a positive width; on a pure algebra AAF item, both algebraic roots may be requested. Read the stem: solutions of the equation means both; possible length means the positive value.

Two consecutive even integers whose product is 48: n(n + 2) = 48, n^2 + 2n − 48 = 0, (n + 8)(n − 6) = 0, n = −8 or n = 6. The pairs are −8, −6 and 6, 8. Creating the product equation is Table 11’s first quadratic skill; factoring it is this section.

Connecting factoring to the other quadratic skills

• Vertex form is not required to solve by factoring, but the midpoint of the two roots is still h.
• If factoring fails quickly, do not stall on an adaptive item: switch to the quadratic formula.
• Inequalities use the same factors, then a sign chart — that is Quadratic Inequalities and Quadratic-Linear Systems.
• Factoring polynomials of degree 3 or 4 is Factoring Cubics and Higher Polynomials; this chapter stays at degree 2.

Keep the zero product property tied to = 0. That one sentence prevents the two highest-frequency errors: treating A · B = k like a zero product, and dividing away an x = 0 root.