Exercise 1 — 25/03/2026 — level ★★☆☆☆
\[ 3x^2 + 6x \]
Answer
\[ 3x(x + 2) \]
Solution
Key Idea
We look for the greatest common factor (GCF) shared by all terms of the polynomial. Here, the GCF is \(3x\).
Identifying the GCF
The coefficients \(3\) and \(6\) have GCD \(3\). The variable \(x\) appears in both terms with exponent at least \(1\). Therefore GCF \(= 3x\).
Factoring Out the GCF
\[ 3x^2 + 6x = 3x \cdot x + 3x \cdot 2 = 3x(x + 2) \]
Check
\[ 3x(x+2) = 3x^2 + 6x \]
Answer
\[ \boxed{3x(x+2)} \]
Exercise 2 — 25/03/2026 — level ★★☆☆☆
\[ 4x^3 - 8x^2 + 12x \]
Answer
\[ 4x(x^2 - 2x + 3) \]
Solution
Identifying the GCF
The coefficients \(4, 8, 12\) have GCD \(4\). The variable \(x\) appears in all terms with exponent at least \(1\). Therefore GCF \(= 4x\).
Factoring Out the GCF
\[ 4x^3 - 8x^2 + 12x = 4x\cdot x^2 - 4x\cdot 2x + 4x\cdot 3 = 4x(x^2 - 2x + 3) \]
Check
\[ 4x(x^2-2x+3) = 4x^3 - 8x^2 + 12x \]
Answer
\[ \boxed{4x(x^2 - 2x + 3)} \]
Exercise 3 — 25/03/2026 — level ★★☆☆☆
\[ 6x^2y - 9xy^2 + 3xy \]
Answer
\[ 3xy(2x - 3y + 1) \]
Solution
Identifying the GCF
The coefficients \(6, 9, 3\) have GCD \(3\). The variable \(x\) appears with exponent at least \(1\), and so does \(y\). Therefore GCF \(= 3xy\).
Factoring Out the GCF
\[ 6x^2y - 9xy^2 + 3xy = 3xy\cdot2x - 3xy\cdot3y + 3xy\cdot1 = 3xy(2x - 3y + 1) \]
Check
\[ 3xy(2x-3y+1) = 6x^2y - 9xy^2 + 3xy \]
Answer
\[ \boxed{3xy(2x - 3y + 1)} \]
Exercise 4 — 25/03/2026 — level ★★☆☆☆
\[ x^2 - 16 \]
Answer
\[ (x-4)(x+4) \]
Solution
Key Idea
We recognize the difference of two squares: \(a^2 - b^2 = (a-b)(a+b)\) with \(a = x\) and \(b = 4\).
Applying the Formula
\[ x^2 - 16 = x^2 - 4^2 = (x-4)(x+4) \]
Check
\[ (x-4)(x+4) = x^2+4x-4x-16 = x^2-16 \]
Answer
\[ \boxed{(x-4)(x+4)} \]
Exercise 5 — 25/03/2026 — level ★★☆☆☆
\[ 9x^2 - 25 \]
Answer
\[ (3x-5)(3x+5) \]
Solution
Key Idea
We recognize the difference of two squares with \(a = 3x\) and \(b = 5\).
Applying the Formula
\[ 9x^2 - 25 = (3x)^2 - 5^2 = (3x-5)(3x+5) \]
Check
\[ (3x-5)(3x+5) = 9x^2+15x-15x-25 = 9x^2-25 \]
Answer
\[ \boxed{(3x-5)(3x+5)} \]
Exercise 6 — 25/03/2026 — level ★★☆☆☆
\[ x^2 + 6x + 9 \]
Answer
\[ (x+3)^2 \]
Solution
Key Idea
We recognize a perfect square trinomial: \(a^2+2ab+b^2 = (a+b)^2\) with \(a=x\) and \(b=3\).
Checking the Structure
First term: \(x^2 = x^2\) \(\checkmark\)
Middle term: \(6x = 2\cdot x\cdot3\) \(\checkmark\)
Last term: \(9 = 3^2\) \(\checkmark\)
Answer
\[ \boxed{(x+3)^2} \]
Exercise 7 — 25/03/2026 — level ★★★☆☆
\[ x^2 + 5x + 6 \]
Answer
\[ (x+2)(x+3) \]
Solution
Key Idea
A monic trinomial of the form \(x^2+bx+c\) factors as \((x+p)(x+q)\) where \(p+q=b\) and \(p\cdot q=c\).
Finding \(p\) and \(q\)
We need two numbers whose product is \(6\) and whose sum is \(5\):
\[ p\cdot q = 6 \qquad p + q = 5 \implies p = 2,\; q = 3 \]
Factored Form
\[ x^2+5x+6 = (x+2)(x+3) \]
Check
\[ (x+2)(x+3) = x^2+3x+2x+6 = x^2+5x+6 \]
Answer
\[ \boxed{(x+2)(x+3)} \]
Exercise 8 — 25/03/2026 — level ★★★☆☆
\[ x^2 - 7x + 12 \]
Answer
\[ (x-3)(x-4) \]
Solution
Finding \(p\) and \(q\)
We need two numbers whose product is \(12\) and whose sum is \(-7\):
\[ p\cdot q = 12 \qquad p+q = -7 \implies p=-3,\; q=-4 \]
Factored Form
\[ x^2-7x+12 = (x-3)(x-4) \]
Check
\[ (x-3)(x-4) = x^2-4x-3x+12 = x^2-7x+12 \]
Answer
\[ \boxed{(x-3)(x-4)} \]
Exercise 9 — 25/03/2026 — level ★★★☆☆
\[ x^2 + x - 6 \]
Answer
\[ (x+3)(x-2) \]
Solution
Finding \(p\) and \(q\)
We need two numbers whose product is \(-6\) and whose sum is \(1\):
\[ p\cdot q = -6 \qquad p+q=1 \implies p=3,\; q=-2 \]
Factored Form
\[ x^2+x-6 = (x+3)(x-2) \]
Check
\[ (x+3)(x-2) = x^2-2x+3x-6 = x^2+x-6 \]
Answer
\[ \boxed{(x+3)(x-2)} \]
Exercise 10 — 25/03/2026 — level ★★★☆☆
\[ 2x^2 + 7x + 3 \]
Answer
\[ (2x+1)(x+3) \]
Solution
Key Idea
For a trinomial \(ax^2+bx+c\) with \(a\neq1\), we use the ac-method: find two numbers whose product is \(ac = 6\) and whose sum is \(b = 7\).
The \(ac\) Product and Factor Search
\[ a\cdot c = 2\cdot3 = 6 \qquad p+q=7 \implies p=1,\; q=6 \]
Splitting the Middle Term
\[ 2x^2+7x+3 = 2x^2+x+6x+3 \]
Factoring by Grouping
\[ = x(2x+1)+3(2x+1) = (2x+1)(x+3) \]
Check
\[ (2x+1)(x+3) = 2x^2+6x+x+3 = 2x^2+7x+3 \]
Answer
\[ \boxed{(2x+1)(x+3)} \]
Exercise 11 — 25/03/2026 — level ★★★☆☆
\[ 3x^2 - 10x + 3 \]
Answer
\[ (3x-1)(x-3) \]
Solution
The \(ac\) Product and Factor Search
\[ a\cdot c = 3\cdot3 = 9 \qquad p+q=-10 \implies p=-1,\; q=-9 \]
Splitting the Middle Term
\[ 3x^2-10x+3 = 3x^2-x-9x+3 \]
Factoring by Grouping
\[ = x(3x-1)-3(3x-1) = (3x-1)(x-3) \]
Check
\[ (3x-1)(x-3) = 3x^2-9x-x+3 = 3x^2-10x+3 \]
Answer
\[ \boxed{(3x-1)(x-3)} \]
Exercise 12 — 25/03/2026 — level ★★★☆☆
\[ 6x^2 + x - 2 \]
Answer
\[ (3x+2)(2x-1) \]
Solution
The \(ac\) Product and Factor Search
\[ a\cdot c = 6\cdot(-2) = -12 \qquad p+q=1 \implies p=4,\; q=-3 \]
Splitting the Middle Term
\[ 6x^2+x-2 = 6x^2+4x-3x-2 \]
Factoring by Grouping
\[ = 2x(3x+2)-(3x+2) = (3x+2)(2x-1) \]
Check
\[ (3x+2)(2x-1) = 6x^2-3x+4x-2 = 6x^2+x-2 \]
Answer
\[ \boxed{(3x+2)(2x-1)} \]
Exercise 13 — 25/03/2026 — level ★★★★☆
\[ x^3 - x \]
Answer
\[ x(x-1)(x+1) \]
Solution
Step 1: Factor Out \(x\)
\[ x^3 - x = x(x^2 - 1) \]
Step 2: Difference of Two Squares
\[ x^2 - 1 = (x-1)(x+1) \]
Complete Factorization
\[ x^3-x = x(x-1)(x+1) \]
Check
\[ x(x-1)(x+1) = x(x^2-1) = x^3-x \]
Answer
\[ \boxed{x(x-1)(x+1)} \]
Exercise 14 — 25/03/2026 — level ★★★★☆
\[ x^3 + 8 \]
Answer
\[ (x+2)(x^2 - 2x + 4) \]
Solution
Key Idea
We recognize the sum of two cubes: \(a^3+b^3=(a+b)(a^2-ab+b^2)\) with \(a=x\) and \(b=2\).
Applying the Formula
\[ x^3+8 = x^3+2^3 = (x+2)(x^2-2x+4) \]
Check
\[ (x+2)(x^2-2x+4) = x^3-2x^2+4x+2x^2-4x+8 = x^3+8 \]
Answer
\[ \boxed{(x+2)(x^2-2x+4)} \]
Exercise 15 — 25/03/2026 — level ★★★★☆
\[ x^3 - 27 \]
Answer
\[ (x-3)(x^2+3x+9) \]
Solution
Key Idea
We recognize the difference of two cubes: \(a^3-b^3=(a-b)(a^2+ab+b^2)\) with \(a=x\) and \(b=3\).
Applying the Formula
\[ x^3-27 = x^3-3^3 = (x-3)(x^2+3x+9) \]
Check
\[ (x-3)(x^2+3x+9) = x^3+3x^2+9x-3x^2-9x-27 = x^3-27 \]
Answer
\[ \boxed{(x-3)(x^2+3x+9)} \]
Exercise 16 — 25/03/2026 — level ★★★★☆
\[ x^3 - x^2 - x + 1 \]
Answer
\[ (x-1)^2(x+1) \]
Solution
Key Idea
We apply factoring by grouping, pairing the terms two by two.
Grouping the Terms
\[ (x^3-x^2)+(-x+1) = x^2(x-1)-(x-1) \]
Factoring Out \((x-1)\)
\[ (x-1)(x^2-1) \]
Further Factoring: Difference of Two Squares
\[ (x-1)(x-1)(x+1) = (x-1)^2(x+1) \]
Check
\[ (x-1)^2(x+1) = (x^2-2x+1)(x+1) = x^3+x^2-2x^2-2x+x+1 = x^3-x^2-x+1 \]
Answer
\[ \boxed{(x-1)^2(x+1)} \]
Exercise 17 — 25/03/2026 — level ★★★★☆
\[ 2x^3 + x^2 - 2x - 1 \]
Answer
\[ (2x+1)(x-1)(x+1) \]
Solution
Factoring by Grouping
\[ (2x^3+x^2)+(-2x-1) = x^2(2x+1)-(2x+1) \]
Factoring Out \((2x+1)\)
\[ (2x+1)(x^2-1) \]
Difference of Two Squares
\[ (2x+1)(x-1)(x+1) \]
Check
\[ (2x+1)(x^2-1) = 2x^3-2x+x^2-1 = 2x^3+x^2-2x-1 \]
Answer
\[ \boxed{(2x+1)(x-1)(x+1)} \]
Exercise 18 — 25/03/2026 — level ★★★★★
\[ x^4 - 1 \]
Answer
\[ (x-1)(x+1)(x^2+1) \]
Solution
Step 1: Difference of Two Squares
\[ x^4-1 = (x^2)^2-1^2 = (x^2-1)(x^2+1) \]
Step 2: Another Difference of Two Squares
\[ x^2-1 = (x-1)(x+1) \]
The factor \(x^2+1\) is irreducible over \(\mathbb{R}\) (discriminant \(-4 < 0\)).
Complete Factorization
\[ x^4-1 = (x-1)(x+1)(x^2+1) \]
Check
\[ (x^2-1)(x^2+1) = x^4+x^2-x^2-1 = x^4-1 \]
Answer
\[ \boxed{(x-1)(x+1)(x^2+1)} \]
Exercise 19 — 25/03/2026 — level ★★★★★
\[ x^4 - 5x^2 + 4 \]
Answer
\[ (x-1)(x+1)(x-2)(x+2) \]
Solution
Key Idea
This is a biquadratic trinomial. We substitute \(t = x^2\) to reduce it to a standard quadratic in \(t\).
Substitution \(t = x^2\)
\[ t^2-5t+4 = 0 \implies (t-1)(t-4) = 0 \implies t=1 \text{ or } t=4 \]
Back-Substitution
\[ t=1 \implies x^2-1=(x-1)(x+1) \]
\[ t=4 \implies x^2-4=(x-2)(x+2) \]
Complete Factorization
\[ x^4-5x^2+4 = (x^2-1)(x^2-4) = (x-1)(x+1)(x-2)(x+2) \]
Check
\[ (x^2-1)(x^2-4) = x^4-4x^2-x^2+4 = x^4-5x^2+4 \]
Answer
\[ \boxed{(x-1)(x+1)(x-2)(x+2)} \]
Exercise 20 — 25/03/2026 — level ★★★★★
\[ x^3 + 3x^2 - 4x - 12 \]
Answer
\[ (x+3)(x-2)(x+2) \]
Solution
Factoring by Grouping
\[ (x^3+3x^2)+(-4x-12) = x^2(x+3)-4(x+3) \]
Factoring Out \((x+3)\)
\[ (x+3)(x^2-4) \]
Difference of Two Squares
\[ x^2-4 = (x-2)(x+2) \]
Complete Factorization
\[ x^3+3x^2-4x-12 = (x+3)(x-2)(x+2) \]
Check
\[ (x+3)(x^2-4) = x^3-4x+3x^2-12 = x^3+3x^2-4x-12 \]
Answer
\[ \boxed{(x+3)(x-2)(x+2)} \]