This page is a self-test. Each problem is stated cleanly, then its complete solution is hidden inside a collapsible callout — try the problem first, then click to reveal. We climb five difficulty levels: L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery.

Every technique used here is built in full on the parent note: the main Factoring topic. Keep it open beside you.

The "always check" ladder

Before any exercise, run this checklist. It is the single most reliable way to never miss a factor.

L1 — Recognition

Problem 1.1

Factor $x^{2} - 25$ .

Recall Solution 1.1

Two terms, both perfect squares: $x^{2} = (x)^{2}$ and $25 = 5^{2}$ . This is a difference of squares $a^{2} - b^{2} = (a - b) (a + b)$ with $a = x, b = 5$ . $x^{2} - 25 = (x - 5) (x + 5)$ Check (expand): $(x - 5) (x + 5) = x^{2} + 5 x - 5 x - 25 = x^{2} - 25.$ ✓

Problem 1.2

Factor $7 x^{3} + 14 x^{2}$ .

Recall Solution 1.2

Two terms. Coefficients: $g cd (7, 14) = 7$ . Lowest power of $x$ present in both terms is $x^{2}$ . So the GCF is $7 x^{2}$ . $\frac{7 x ^{3}}{7 x ^{2}} = x, \frac{14 x ^{2}}{7 x ^{2}} = 2$ $7 x^{3} + 14 x^{2} = 7 x^{2} (x + 2)$ Check: $7 x^{2} (x + 2) = 7 x^{3} + 14 x^{2} .$ ✓

Problem 1.3

Factor $y^{2} + 10 y + 25$ .

Recall Solution 1.3

Three terms. The outer terms are squares: $y^{2} = (y)^{2}$ , $25 = 5^{2}$ . Test the middle term against the perfect-square pattern $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ with $a = y, b = 5$ : $2 ab = 2 (y) (5) = 10 y ✓ matches the middle term.$ $y^{2} + 10 y + 25 = (y + 5)^{2}$

L2 — Application

Problem 2.1

Factor completely: $2 x^{2} + 8 x + 6$ .

Recall Solution 2.1

Step 1 — GCF first. $g cd (2, 8, 6) = 2$ , no variable common to all as a guaranteed lowest power beyond $x^{0}$ , so GCF $= 2$ : $2 x^{2} + 8 x + 6 = 2 (x^{2} + 4 x + 3)$ Step 2 — factor the trinomial. Find two numbers multiplying to $3$ and adding to $4$ : that is $1$ and $3$ . $x^{2} + 4 x + 3 = (x + 1) (x + 3)$ $2 x^{2} + 8 x + 6 = 2 (x + 1) (x + 3)$ Check: $2 (x + 1) (x + 3) = 2 (x^{2} + 4 x + 3) = 2 x^{2} + 8 x + 6.$ ✓

Problem 2.2

Factor $9 m^{2} - 30 m + 25$ .

Recall Solution 2.2

Outer terms are squares: $9 m^{2} = (3 m)^{2}$ , $25 = 5^{2}$ . Test perfect square $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ with $a = 3 m, b = 5$ : $2 ab = 2 (3 m) (5) = 30 m ✓$ The sign is $-$ , so use $(a - b)^{2}$ : $9 m^{2} - 30 m + 25 = (3 m - 5)^{2}$

Problem 2.3

Factor $27 t^{3} - 8$ .

Recall Solution 2.3

Two terms, both cubes: $27 t^{3} = (3 t)^{3}$ , $8 = 2^{3}$ . Difference of cubes $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ with $a = 3 t, b = 2$ : $27 t^{3} - 8 = (3 t - 2) ((3 t)^{2} + (3 t) (2) + 2^{2}) = (3 t - 2) (9 t^{2} + 6 t + 4)$ Check: $(3 t - 2) (9 t^{2} + 6 t + 4) = 27 t^{3} + 18 t^{2} + 12 t - 18 t^{2} - 12 t - 8 = 27 t^{3} - 8.$ ✓

L3 — Analysis

Problem 3.1

Factor $x^{3} + 5 x^{2} + 3 x + 15$ .

Recall Solution 3.1

Four terms, no factor common to all four. Try grouping in the given order: $(x^{3} + 5 x^{2}) + (3 x + 15) = x^{2} (x + 5) + 3 (x + 5)$ Both groups now share the binomial $(x + 5)$ — pull it out: $= (x + 5) (x^{2} + 3)$ $x^{2} + 3$ is a sum, not factorable over the reals, so we stop. $x^{3} + 5 x^{2} + 3 x + 15 = (x + 5) (x^{2} + 3)$

Problem 3.2

Factor $6 x y + 4 x - 15 y - 10$ .

Recall Solution 3.2

Four terms. Group the first two and last two: $(6 x y + 4 x) + (- 15 y - 10) = 2 x (3 y + 2) - 5 (3 y + 2)$ Notice we factored $- 5$ (not $+ 5$ ) out of the second group so that $(3 y + 2)$ appears — matching the first group. Now extract the common binomial: $= (3 y + 2) (2 x - 5)$ Check: $(3 y + 2) (2 x - 5) = 6 x y - 15 y + 4 x - 10 = 6 x y + 4 x - 15 y - 10.$ ✓

Problem 3.3

Factor $ab + 3 c + 3 b + a c$ .

Recall Solution 3.3

Group as given: $(ab + 3 c) + (3 b + a c)$ — no common binomial. Reorder so terms sharing a factor sit together: $ab + a c + 3 b + 3 c = a (b + c) + 3 (b + c) = (b + c) (a + 3)$ Why reorder? Grouping only works when each pair leaves the same binomial. Rearranging is a legal move (addition commutes) that engineers that match.

L4 — Synthesis

Problem 4.1

Factor completely: $3 x^{4} - 48$ .

Recall Solution 4.1

Step 1 — GCF. $g cd (3, 48) = 3$ : $3 x^{4} - 48 = 3 (x^{4} - 16)$ Step 2 — difference of squares. $x^{4} = (x^{2})^{2}$ , $16 = 4^{2}$ : $= 3 (x^{2} - 4) (x^{2} + 4)$ Step 3 — factor again. $x^{2} - 4 = (x - 2) (x + 2)$ (another difference of squares). $x^{2} + 4$ is a sum of squares, prime over the reals. $3 x^{4} - 48 = 3 (x - 2) (x + 2) (x^{2} + 4)$

Problem 4.2

Factor completely: $2 a^{3} - 2 a^{2} - 24 a$ .

Recall Solution 4.2

Step 1 — GCF $= 2 a$ : $2 a^{3} - 2 a^{2} - 24 a = 2 a (a^{2} - a - 12)$ Step 2 — factor the trinomial. Two numbers multiplying to $- 12$ , adding to $- 1$ : that is $- 4$ and $3$ . $a^{2} - a - 12 = (a - 4) (a + 3)$ $2 a^{3} - 2 a^{2} - 24 a = 2 a (a - 4) (a + 3)$ Check: $2 a (a - 4) (a + 3) = 2 a (a^{2} - a - 12) = 2 a^{3} - 2 a^{2} - 24 a .$ ✓

Problem 4.3

Factor $x^{6} - 64$ .

Recall Solution 4.3

Here $x^{6} = (x^{2})^{3} = (x^{3})^{2}$ and $64 = 4^{3} = 8^{2}$ , so we could treat it as a difference of squares or cubes. Squares first gives the fullest split: $x^{6} - 64 = (x^{3})^{2} - 8^{2} = (x^{3} - 8) (x^{3} + 8)$ Now each factor is a cube. Difference of cubes ( $a = x, b = 2$ ) and sum of cubes ( $a = x, b = 2$ ): $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4), x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$ $x^{6} - 64 = (x - 2) (x + 2) (x^{2} + 2 x + 4) (x^{2} - 2 x + 4)$ Both quadratics are prime over the reals (their discriminants are negative). See the geometric picture below.

Figure — Factoring — common factor extraction, grouping, using identities

L5 — Mastery

Problem 5.1

Factor completely over the reals, then over the complex numbers: $x^{4} + 5 x^{2} + 4$ .

Recall Solution 5.1

Treat it as a quadratic in $u = x^{2}$ : $u^{2} + 5 u + 4 = (u + 1) (u + 4)$ . Substitute back: $x^{4} + 5 x^{2} + 4 = (x^{2} + 1) (x^{2} + 4)$ Over the reals, both are sums of squares — prime. This is the complete real factorization. $x^{4} + 5 x^{2} + 4 = (x^{2} + 1) (x^{2} + 4) (over R)$ Over the complex numbers (Complex Numbers, Fundamental Theorem of Algebra), each splits: $x^{2} + 1 = (x - i) (x + i), x^{2} + 4 = (x - 2 i) (x + 2 i)$ $x^{4} + 5 x^{2} + 4 = (x - i) (x + i) (x - 2 i) (x + 2 i)$ The degree-4 polynomial has exactly 4 complex roots ( $\pm i, \pm 2 i$ ) — exactly as the Fundamental Theorem of Algebra promises.

Problem 5.2

Factor $2 x^{3} + 3 x^{2} - 8 x - 12$ completely, using grouping.

Recall Solution 5.2

Four terms. Group: $(2 x^{3} + 3 x^{2}) + (- 8 x - 12) = x^{2} (2 x + 3) - 4 (2 x + 3) = (2 x + 3) (x^{2} - 4)$ Now $x^{2} - 4$ is a difference of squares: $= (2 x + 3) (x - 2) (x + 2)$ $2 x^{3} + 3 x^{2} - 8 x - 12 = (2 x + 3) (x - 2) (x + 2)$ The roots are $x = - \frac{3}{2}, 2, - 2$ (each factor set to zero via the Zero Product Property). Check: $(2 x + 3) (x^{2} - 4) = 2 x^{3} - 8 x + 3 x^{2} - 12 = 2 x^{3} + 3 x^{2} - 8 x - 12.$ ✓

Problem 5.3

Factor $x^{3} - 3 x^{2} - 4 x + 12$ completely.

Recall Solution 5.3

Group: $(x^{3} - 3 x^{2}) + (- 4 x + 12) = x^{2} (x - 3) - 4 (x - 3) = (x - 3) (x^{2} - 4)$ Difference of squares again: $= (x - 3) (x - 2) (x + 2)$ $x^{3} - 3 x^{2} - 4 x + 12 = (x - 3) (x - 2) (x + 2)$ Roots $x = 3, 2, - 2$ . (You could also have found these first with the Rational Root Theorem and confirmed by Polynomial Long Division — grouping just did it faster.)

Problem 5.4

Is $x^{2} + x + 1$ factorable over the reals? Justify. Then factor over the complex numbers.

Recall Solution 5.4

It is a trinomial with no obvious integer split (no two integers multiply to $1$ and add to $1$ ). Use the discriminant from the Quadratic Formula — the number under the square root, $b^{2} - 4 a c$ : $b^{2} - 4 a c = 1^{2} - 4 (1) (1) = 1 - 4 = - 3 < 0$ A negative discriminant means there are no real roots, so $x^{2} + x + 1$ is prime (irreducible) over the reals. Over $C$ , the roots are $x = \frac{- 1 \pm - 3}{2} = \frac{- 1 \pm i 3}{2}$ , giving $x^{2} + x + 1 = (x - \frac{- 1 + i 3}{2}) (x - \frac{- 1 - i 3}{2}) .$

Level checkpoint (reveal answers)

What identity fits

x^{2} - 25

Difference of squares

\to (x - 5) (x + 5)

What must you always do before any other factoring?

Pull out the GCF.

Difference of cubes

a^{3} - b^{3}

factors as?

(a - b) (a^{2} + ab + b^{2})

— middle sign opposite the binomial.

When does the first grouping fail but factoring still work?

When terms need reordering so pairs share a common binomial.

Fully factor

x^{4} - 16

(x - 2) (x + 2) (x^{2} + 4)

How do you know

x^{2} + x + 1

is prime over

R

Discriminant

b^{2} - 4 a c = - 3 < 0

, so no real roots.