Using the AC method: $(x + 2)(x + 4)$

Using the difference of squares: $(5x + 4)(5x - 4)$

Using perfect square trinomials: $(x + 5)^2$

Using the difference of cubes: $(2x - 3)(4x^2 + 6x + 9)$

Factor to $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$.

Expand to $x^2 + 2x - 35 = 0$, factor to $(x + 7)(x - 5) = 0$, so $x = 5$ (discard -7).

Using the AC method, find factors of -6 that add to 5 (6 and -1). Rewrite as $2x^2 + 6x - x - 3$. Factor by grouping: $2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)$.

Recognize this as a difference of squares: $(3x)^2 - 7^2$. Apply the formula: $(3x - 7)(3x + 7)$.

Factor the quadratic: $(x - 5)(x + 1) = 0$. Set each factor to zero: $x - 5 = 0$ or $x + 1 = 0$. Solve for $x$: $x = 5$ or $x = -1$.

Factor the area: $x^2 + 7x + 10 = (x + 2)(x + 5)$. Since Area = Length * Width, the width is $x + 2$.

Rewriting a polynomial as a product of its factors.

$a^2 - b^2 = (a + b)(a - b)$

A trinomial that can be factored into $(a + b)^2$ or $(a - b)^2$.

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

If the product of factors is zero, at least one of the factors must be zero.

An equation in the form $ax^2 + bx + c = 0$ where $a \neq 0$.

A factoring technique where you find two numbers that multiply to ac and add to b in a quadratic expression $ax^2 + bx + c$.

A factoring method where terms with common factors are grouped, and the greatest common factor (GCF) is pulled out from each group.

Rewriting the quadratic as $(x + p)^2 + q$.