Expanding and Factoring Expressions: Additional Practice (Set 2)

The middle products $5x$ and $-5x$ cancel.

$(x - 5)(x + 5) = x^2 + 5x - 5x - 25 = x^2 - 25$

Combine the middle products $2x$ and $-7x$.

$(x - 7)(x + 2) = x^2 + 2x - 7x - 14 = x^2 - 5x - 14$

The middle products are $x$ and $-6x$, which combine to $-5x$.

$(x - 3)(2x + 1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$

Expand the product, where the middle terms cancel, then add $25$.

$(x + 5)(x - 5) + 25 = x^2 - 25 + 25 = x^2$

Expand each; three give $6x + 9$ but one does not.

$3(2x + 9) = 6x + 27 \neq 6x + 9$

So $3(2x + 9)$ is the expression that is not equivalent.

The middle products $-xy$ and $xy$ cancel.

$(x + y)(x - y) = x^2 - xy + xy - y^2 = x^2 - y^2$

The lowest power of $x$ common to both terms is $x^2$.

$x^3 + x^2 = x^2(x + 1)$

Multiply every term of $x - 3$ by every term of the trinomial, then combine.

$(x - 3)(x^2 + 3x + 9) = x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 = x^3 - 27$

The $x^2$ and $x$ terms cancel.

The middle products $6x$ and $-6x$ cancel; $3x \cdot 3x = 9x^2$.

$(3x - 2)(3x + 2) = 9x^2 + 6x - 6x - 4 = 9x^2 - 4$

The greatest common factor of $18x$ and $24$ is $6$; smaller factors like $2$ or $3$ leave more to pull out.

$18x + 24 = 6(3x + 4)$

Multiply every term of $x + 2$ by every term of the trinomial, then combine.

$(x + 2)(x^2 - 2x + 4) = x^3 - 2x^2 + 4x + 2x^2 - 4x + 8 = x^3 + 8$

The $x^2$ and $x$ terms cancel.

The middle products $4x$ and $-4x$ cancel.

$(x - 4)(x + 4) = x^2 + 4x - 4x - 16 = x^2 - 16$