Factoring

A1.I.1Factor out GCF01/05

Factor out the greatest common factor.

Factor this expression

12 x^{3} + 18 x^{2} - 30 x

Step01of 05

Factoring is the reverse of distributing. Where distribution expanded $3 (x + 4)$ into $3 x + 12$ , factoring starts with the expanded form and pulls a common chunk out front.

Always look for the GCF first. Before any other factoring technique, check what every term shares.

A1.I.1 — Factor out the greatest common factor.

Factor this expression

12 x^{3} + 18 x^{2} - 30 x

Step 01 of 05

Factoring is the reverse of distributing. Where distribution expanded $3 (x + 4)$ into $3 x + 12$ , factoring starts with the expanded form and pulls a common chunk out front.

Always look for the GCF first. Before any other factoring technique, check what every term shares.

Step 02 of 05

Step 1 — find the GCF of the coefficients.

12 = 2^{2} \cdot 318 = 2 \cdot 3^{2} 30 = 2 \cdot 3 \cdot 5

All three share $2 \cdot 3 = 6$ . So the numerical GCF is $6$ .

Step 03 of 05

Step 2 — find the GCF of the variable parts. Every term has at least $x^{1}$ (the powers are $3, 2, 1$ ). Take the LOWEST power present.

variable GCF = x^{1} = x

Combine: full GCF = $6 x$ .

Step 04 of 05

Step 3 — divide every term by the GCF and write what's left inside parens.

\frac{12 x ^{3}}{6 x} = 2 x^{2} \frac{18 x ^{2}}{6 x} = 3 x \frac{- 30 x}{6 x} = - 5

12 x^{3} + 18 x^{2} - 30 x = 6 x (2 x^{2} + 3 x - 5)

Step 05 of 05

Step 4 — verify by re-distributing.

6 x (2 x^{2} + 3 x - 5) = 12 x^{3} + 18 x^{2} - 30 x ✓

Matches the original. The expression inside the parens may itself factor further (here $2 x^{2} + 3 x - 5$ factors as $(2 x + 5) (x - 1)$ ) — that's the next skills.

Key insight

GCF is the always-first move. Even when the expression doesn't fully factor, pulling out the GCF makes whatever's left smaller and easier to handle.

A1.I.2 — Factor x² + bx + c.

Factor this trinomial

x^{2} + 7 x + 12

Step 01 of 05

When the leading coefficient is $1$ , the factored form looks like:

x^{2} + b x + c = (x + p) (x + q)

FOIL the right side: $p$ and $q$ must satisfy $p + q = b$ AND $p \cdot q = c$ . Find two numbers that add to b and multiply to $c$ .

Step 02 of 05

For our trinomial $x^{2} + 7 x + 12$ : need $p + q = 7$ and $p \cdot q = 12$ .

List factor pairs of $12$ :

1, 12	→sum = 13
2, 6	→sum = 8
3, 4	→sum = 7 ✓

Found it: $p = 3, q = 4$ .

Step 03 of 05

Write the factored form.

x^{2} + 7 x + 12 = (x + 3) (x + 4)

Verify by FOIL: $(x + 3) (x + 4) = x^{2} + 4 x + 3 x + 12 = x^{2} + 7 x + 12 ✓$ .

Step 04 of 05

Sign rules. Watching the signs of b and $c$ tells you which signs $p$ and $q$ need.

$c > 0, b > 0$	→both positive
$c > 0, b < 0$	→both negative
$c < 0$	→opposite signs (the larger absolute value matches sign of b)

Example: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ — both negative because $c > 0$ and $b < 0$ .

Step 05 of 05

If no integer pair works, the trinomial doesn't factor over the integers — it's "prime" (irreducible). That's not a mistake; it's a valid answer.

Example: $x^{2} + 5 x + 7$ . No two integers multiply to $7$ (only $1, 7$ ) AND add to $5$ . Can't factor over integers — would need the quadratic formula.

Key insight

Two numbers that ADD to b and MULTIPLY to

c

. List factor pairs of

c

, scan their sums, take the pair whose sum is b. If none works, it's prime.

A1.I.3 — Factor ax² + bx + c (the AC method).

Factor this trinomial

6 x^{2} + 11 x + 4

Step 01 of 05

When a isn't $1$ , the simple "two numbers" trick from the previous skill doesn't quite work. The AC method(also called factoring by grouping) handles it cleanly.

Step 1 — compute $a \cdot c$ .

a \cdot c = 6 \cdot 4 = 24

Step 02 of 05

Step 2 — find two numbers that multiply to $a c$ and add to b. Same hunt as before, but the product target is $a c$ instead of $c$ .

Pairs that multiply to $24$ :

1, 24	→sum = 25
2, 12	→sum = 14
3, 8	→sum = 11 ✓
4, 6	→sum = 10

The winning pair: $3, 8$ .

Step 03 of 05

Step 3 — split the middle term using those two numbers. Replace $11 x$ with $3 x + 8 x$ .

6 x^{2} + 11 x + 4 = 6 x^{2} + 3 x + 8 x + 4

Now there are FOUR terms — perfect for grouping.

Step 04 of 05

Step 4 — group in pairs, GCF each pair.

(6 x^{2} + 3 x) + (8 x + 4)

= 3 x (2 x + 1) + 4 (2 x + 1)

Both groups now share the binomial $(2 x + 1)$ . That match is the signal you split correctly.

Step 05 of 05

Step 5 — factor out the common binomial.

= (2 x + 1) (3 x + 4)

Verify by FOIL: $(2 x + 1) (3 x + 4) = 6 x^{2} + 8 x + 3 x + 4 = 6 x^{2} + 11 x + 4 ✓$ .

Key insight

AC method in five beats: compute

a c

, find a pair that multiplies to

a c

and adds to b, split the middle term into those two pieces, group, factor each pair, and the shared binomial pops out.

A1.I.4 — Factor a difference of squares.

The pattern to recognize

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

Step 01 of 05

A difference of squares is two perfect-square terms separated by a MINUS sign. It factors instantly into a sum and a difference of the square roots.

Has the pattern	→ $x^{2} - 9, 25 x^{2} - 16, x^{4} - 1$
Does NOT	→ $x^{2} + 9$ (sum, not difference)
Does NOT	→ $x^{2} - 7$ (7 isn't a perfect square — leave alone)

Step 02 of 05

Worked example.

x^{2} - 49

Identify the squares: $x^{2} = (x)^{2}$ , $49 = 7^{2}$ . So $a = x$ and $b = 7$ .

x^{2} - 49 = (x - 7) (x + 7)

Step 03 of 05

With a coefficient on $x^{2}$ .

25 x^{2} - 16

Squares: $25 x^{2} = (5 x)^{2}$ , $16 = 4^{2}$ . So $a = 5 x$ , $b = 4$ .

25 x^{2} - 16 = (5 x - 4) (5 x + 4)

Verify: $(5 x)^{2} - 4^{2} = 25 x^{2} - 16 ✓$ .

Step 04 of 05

Pull out a GCF first if there is one. Always check.

50 x^{2} - 8 = 2 (25 x^{2} - 4) = 2 (5 x - 2) (5 x + 2)

Without the GCF step you'd be stuck — $50 x^{2} - 8$ isn't a clean difference of squares. After pulling out $2$ , what's left IS.

Step 05 of 05

Sometimes you can factor twice. Difference of squares may itself contain a difference of squares.

x^{4} - 16 = (x^{2} - 4) (x^{2} + 4)

The first factor $x^{2} - 4$ is ANOTHER difference of squares:

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

The $x^{2} + 4$ is a SUM of squares — doesn't factor over the reals. Leave it.

Key insight

Two perfect squares with a minus → factors instantly as

(first - second) (first + second)

. Sum of squares does NOT factor over the reals — that's the asymmetry worth memorizing.

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