🧮 Algebra Factoring And Identities (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Difference of Squares · Easy · source: Original (AMC-style)
Factor $x^2 - 9$.
A) $(x-3)^2$
B) $(x+3)^2$
C) $(x-3)(x+3)$
D) $(x-9)(x+1)$
E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of squares: $x^2 - 9 = x^2 - 3^2 = (x-3)(x+3)$.
2.
Tags: Perfect Square · Easy · source: Original (AMC-style)
Factor $x^2 + 6x + 9$.
A) $(x+3)^2$
B) $(x-3)^2$
C) $(x+3)(x-3)$
D) $(x+6)(x+1)$
E) Cannot be factored
Answer & Solution
Answer: A
This is a perfect square: $x^2 + 6x + 9 = (x+3)^2$.
3.
Tags: Difference of Cubes · Easy · source: Original (AMC-style)
Factor $x^3 - 8$.
A) $(x-2)^3$
B) $(x+2)^3$
C) $(x-2)(x^2+2x+4)$
D) $(x+2)(x^2-2x+4)$
E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of cubes: $x^3 - 8 = x^3 - 2^3 = (x-2)(x^2+2x+4)$.
4.
Tags: Sum of Cubes · Easy · source: Original (AMC-style)
Factor $x^3 + 27$.
A) $(x+3)^3$
B) $(x-3)^3$
C) $(x+3)(x^2-3x+9)$
D) $(x-3)(x^2+3x+9)$
E) Cannot be factored
Answer & Solution
Answer: C
This is a sum of cubes: $x^3 + 27 = x^3 + 3^3 = (x+3)(x^2-3x+9)$.
5.
Tags: GCF Factoring · Medium · source: Original (AMC-style)
Factor $6x^2 + 9x$.
A) $3x(2x+3)$
B) $6x(x+3)$
C) $9x(2x+1)$
D) $3(2x^2+3x)$
E) Cannot be factored
Answer & Solution
Answer: A
Factor out the GCF: $6x^2 + 9x = 3x(2x + 3)$.
6.
Tags: Quadratic Factoring · Medium · source: Original (AMC-style)
Factor $x^2 - 5x + 6$.
A) $(x-2)(x-3)$
B) $(x+2)(x+3)$
C) $(x-1)(x-6)$
D) $(x+1)(x+6)$
E) Cannot be factored
Answer & Solution
Answer: A
We need two numbers that multiply to 6 and add to -5. These are -2 and -3: $x^2 - 5x + 6 = (x-2)(x-3)$.
7.
Tags: Quadratic Factoring · Medium · source: Original (AMC-style)
Factor $2x^2 + 7x + 3$.
A) $(2x+1)(x+3)$
B) $(2x+3)(x+1)$
C) $(x+1)(2x+3)$
D) $(x+3)(2x+1)$
E) Cannot be factored
Answer & Solution
Answer: A
We need two numbers that multiply to $2 \cdot 3 = 6$ and add to 7. These are 1 and 6. So $2x^2 + 7x + 3 = (2x+1)(x+3)$.
8.
Tags: Perfect Square · Medium · source: Original (AMC-style)
Factor $4x^2 - 12x + 9$.
A) $(2x-3)^2$
B) $(2x+3)^2$
C) $(4x-3)(x-3)$
D) $(4x+3)(x+3)$
E) Cannot be factored
Answer & Solution
Answer: A
This is a perfect square: $4x^2 - 12x + 9 = (2x)^2 - 2(2x)(3) + 3^2 = (2x-3)^2$.
9.
Tags: Difference of Squares · Hard · source: Original (AMC-style)
Factor $x^4 - 16$.
A) $(x^2-4)^2$
B) $(x^2+4)^2$
C) $(x^2-4)(x^2+4)$
D) $(x-2)^2(x+2)^2$
E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of squares: $x^4 - 16 = (x^2)^2 - 4^2 = (x^2-4)(x^2+4)$. We can factor further: $(x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)$.
10.
Tags: Grouping · Hard · source: Original (AMC-style)
Factor $x^3 + x^2 + x + 1$.
A) $(x+1)^3$
B) $(x^2+1)(x+1)$
C) $(x^2+1)(x-1)$
D) $(x+1)(x^2-x+1)$
E) Cannot be factored
Answer & Solution
Answer: B
Group the terms: $x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1)$.
11.
Tags: Complex Factoring · Hard · source: Original (AMC-style)
Factor $x^4 + 4$.
A) $(x^2+2)^2$
B) $(x^2-2)^2$
C) $(x^2+2x+2)(x^2-2x+2)$
D) $(x^2+2)(x^2-2)$
E) Cannot be factored
Answer & Solution
Answer: C
Add and subtract $4x^2$: $x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2+2)^2 - (2x)^2 = (x^2+2x+2)(x^2-2x+2)$.
12.
Tags: Identity Application · Hard · source: Original (AMC-style)
If $a + b = 5$ and $ab = 6$, what is $a^2 + b^2$?
A) $13$
B) $17$
C) $19$
D) $25$
E) $37$
Answer & Solution
Answer: A
Using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $5^2 = a^2 + 2(6) + b^2$, so $25 = a^2 + 12 + b^2$. Therefore $a^2 + b^2 = 13$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | C | A | C | C | A | A | A | A | C | B | C | A |