Factoring with GCF First

Factor completely: $3x^3 - 12x$

Find the GCF: GCF of $3x^3$ and $12x$ is $3x$ = $3x(x^2 - 4)$

Check remaining factor: $x^2 - 4$ is a difference of squares: $a^2 - b^2$ where $a = x$, $b = 2$ = Can be factored further

Factor the difference of squares: $x^2 - 4 = (x + 2)(x - 2)$ = $(x + 2)(x - 2)$

Write complete factorization: Combine GCF with factored form = $3x(x + 2)(x - 2)$

Answer: $3x(x + 2)(x - 2)$

Multiple Factoring Techniques

Factor completely: $2x^4 - 32$

Find the GCF: GCF of $2x^4$ and $32$ is $2$ = $2(x^4 - 16)$

Recognize pattern in remaining factor: $x^4 - 16 = (x^2)^2 - 4^2$ is a difference of squares = Apply $a^2 - b^2 = (a+b)(a-b)$

Factor the difference of squares: $(x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$ = $2(x^2 + 4)(x^2 - 4)$

Check each factor: $x^2 + 4$ cannot be factored (sum of squares). $x^2 - 4$ is another difference of squares! = $x^2 - 4 = (x + 2)(x - 2)$

Write complete factorization: Combine all factors = $2(x^2 + 4)(x + 2)(x - 2)$

Answer: $2(x^2 + 4)(x + 2)(x - 2)$

Trinomial with GCF

Factor completely: $4x^2 + 20x + 24$

Find the GCF: GCF of $4$, $20$, and $24$ is $4$ = $4(x^2 + 5x + 6)$

Factor the trinomial: Find two numbers that multiply to $6$ and add to $5$: $2$ and $3$ = $(x + 2)(x + 3)$

Write complete factorization: Combine GCF with trinomial factors = $4(x + 2)(x + 3)$

Verify: Check each factor: $(x + 2)$ and $(x + 3)$ are prime (cannot factor further) = Completely factored

Answer: $4(x + 2)(x + 3)$

Grouping After GCF

Factor completely: $6x^3 + 9x^2 - 2x - 3$

Check for GCF: GCF of all terms is $1$ (no common factor) = Proceed to grouping

Group terms in pairs: $(6x^3 + 9x^2) + (-2x - 3)$ = Two groups

Factor GCF from each group: $3x^2(2x + 3) - 1(2x + 3)$ = Common binomial: $(2x + 3)$

Factor out common binomial: $(2x + 3)(3x^2 - 1)$ = $(2x + 3)(3x^2 - 1)$

Check remaining factors: $3x^2 - 1$ is not a perfect square difference (would need $\sqrt{3}x$) = Cannot factor further with integers

Answer: $(2x + 3)(3x^2 - 1)$

Sum of Cubes with GCF

Factor completely: $2x^3 + 16$

Find the GCF: GCF of $2x^3$ and $16$ is $2$ = $2(x^3 + 8)$

Recognize the pattern: $x^3 + 8 = x^3 + 2^3$ is a sum of cubes = Apply $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Apply sum of cubes formula: With $a = x$ and $b = 2$: $(x + 2)(x^2 - 2x + 4)$ = $(x + 2)(x^2 - 2x + 4)$

Check trinomial factor: $x^2 - 2x + 4$ has discriminant $4 - 16 = -12 < 0$ = Cannot factor with real numbers

Write complete factorization: Combine all factors = $2(x + 2)(x^2 - 2x + 4)$

Answer: $2(x + 2)(x^2 - 2x + 4)$

Mistake: Forgetting to factor out the GCF first

Why: Starting with other techniques when there's a GCF makes the problem harder and can lead to incomplete factoring.

Correct: Always check for GCF before any other factoring technique. $6x^2 - 24 = 6(x^2 - 4) = 6(x+2)(x-2)$

Mistake: Stopping too early

Why: After one factoring step, students often forget to check if factors can be factored further.

Correct: Always examine each factor. $x^4 - 1 = (x^2+1)(x^2-1)$ but $(x^2-1) = (x+1)(x-1)$, so the complete answer is $(x^2+1)(x+1)(x-1)$

Mistake: Trying to factor a sum of squares

Why: $a^2 + b^2$ cannot be factored using real numbers (no real factors).

Correct: Only difference of squares factors: $x^2 - 9 = (x+3)(x-3)$, but $x^2 + 9$ is already prime.

Mistake: Incorrect sign in grouping

Why: When factoring a negative from a group, students forget to change signs inside.

Correct: $-2x - 6 = -2(x + 3)$, not $-2(x - 3)$. Check by distributing back!

Engineering: Projectile Motion

Engineers use complete factorization to find when and where objects land.

A ball's height is modeled by $h = -16t^2 + 48t$. Factor to find when it hits the ground: $-16t(t - 3) = 0$, so $t = 0$ or $t = 3$ seconds.

Computer Science: Algorithm Optimization

Programmers factor expressions to simplify calculations and improve efficiency.

Instead of computing $n^3 - n$ separately, factor to $n(n-1)(n+1)$ - the product of three consecutive integers!

Q: How do I know when a polynomial is completely factored?

A: A polynomial is completely factored when: (1) the GCF is factored out, (2) no factor is a difference of squares that can be factored, (3) no trinomial factor can be factored further, and (4) all remaining factors are prime (linear or irreducible quadratic).

Q: What's the difference between 'factoring' and 'factoring completely'?

A: Regular factoring might stop after one step. Factoring completely means continuing until no factor can be broken down further. For example, $x^4 - 16 = (x^2 + 4)(x^2 - 4)$ is factored, but $(x^2 + 4)(x + 2)(x - 2)$ is factored completely.

Q: Why can't we factor $x^2 + 4$?

A: The sum of two squares, $a^2 + b^2$, has no real factors. Unlike $a^2 - b^2 = (a+b)(a-b)$, there's no similar formula for sums. The expression $x^2 + 4$ is called 'prime' or 'irreducible' over the real numbers.