Expanding and Factoring Expressions: Additional Practice (Set 1)

Expand $(3x - 1)(3x - 1)$; the two cross products are each $-3x$.

$(3x - 1)^2 = 9x^2 - 3x - 3x + 1 = 9x^2 - 6x + 1$

Expand each; three return $8x^2 + 12x$ but one does not.

$2(4x^2 + 6) = 8x^2 + 12 \neq 8x^2 + 12x$

So $2(4x^2 + 6)$ is not equivalent.

The coefficients $9$ and $12$ share $3$, and the lowest power of $x$ is $x^2$, so the GCF is $3x^2$.

$9x^3 - 12x^2 = 3x^2(3x - 4)$

Multiply the binomials, then negate every term.

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

The leading minus flips the sign of every term.

The middle products are $-12x$ and $2x$, combining to $-10x$.

$(4x + 1)(2x - 3) = 8x^2 - 12x + 2x - 3 = 8x^2 - 10x - 3$

The coefficients $16$ and $24$ share $8$, and both terms have $x$, so the GCF is $8x$; the other choices leave a shared factor inside.

$16x^2 + 24x = 8x(2x + 3)$

The coefficients $25$ and $10$ share $5$, and both terms have $x$, so the GCF is $5x$.

$25x^2 - 10x = 5x(5x - 2)$

Expand both squares, then subtract, distributing the minus.

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

The middle products $15x$ and $-2x$ combine to $13x$.

$(5x - 2)(x + 3) = 5x^2 + 15x - 2x - 6 = 5x^2 + 13x - 6$

The coefficients $6$ and $8$ share $2$, and both terms have $x$, so the GCF is $2x$.

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

The middle products $-15x$ and $2x$ combine to $-13x$.

$(3x + 2)(x - 5) = 3x^2 - 15x + 2x - 10 = 3x^2 - 13x - 10$

Expand the product, then multiply each term by $-2$.

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