Expanding and Factoring Expressions: Practice

Multiply each term of $x + 2$ by all three terms, then combine like terms.

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

The middle products are $8x$ and $-3x$, which combine to $5x$.

$(2x - 1)(3x + 4) = 6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4$

The coefficients $6$ and $9$ share $3$, and the lowest power of $x$ present is $x^2$, so the GCF is $3x^2$.

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

The coefficients $15$ and $20$ share $5$, and both terms have $x$, so the GCF is $5x$.

$15x^2 - 20x = 5x(3x - 4)$

Expand $(2x + 3)(2x + 3)$; the two cross products are each $6x$.

$(2x + 3)^2 = 4x^2 + 6x + 6x + 9 = 4x^2 + 12x + 9$

Multiply the two binomials, then apply the leading minus to every term.

$-(x - 2)(x + 5) = -(x^2 + 3x - 10) = -x^2 - 3x + 10$

The leading minus flips the sign of every term.

Expand the product, then multiply through by $2$.

$2(x + 3)(x - 1) = 2(x^2 + 2x - 3) = 2x^2 + 4x - 6$

Expand both squares, then subtract, distributing the minus.

$(x + 4)^2 - (x - 4)^2 = (x^2 + 8x + 16) - (x^2 - 8x + 16) = 16x$

The coefficients $14$ and $21$ share $7$, and both terms have $x$, so the GCF is $7x$.

$14x^2 + 21x = 7x(2x + 3)$

The middle products cancel; $2x \cdot 2x = 4x^2$.

$(2x + 1)(2x - 1) = 4x^2 - 2x + 2x - 1 = 4x^2 - 1$

Expand both products, then subtract the second, flipping its signs.

$(x + 3)(x - 2) - (x + 1)(x - 4) = (x^2 + x - 6) - (x^2 - 3x - 4) = 4x - 2$

Combine the middle products $4x$ and $-6x$.

$(x - 6)(x + 4) = x^2 + 4x - 6x - 24 = x^2 - 2x - 24$