Difference of Squares: Practice

Take the root of each term: $\sqrt{9x^2} = 3x$ and $\sqrt{16} = 4$.

$9x^2 - 16 = (3x)^2 - 4^2 = (3x + 4)(3x - 4)$

The root of $9x^2$ is $3x$, not $9x$.

The roots are $\sqrt{4x^2} = 2x$ and $\sqrt{9} = 3$.

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

The root of $4x^2$ is $2x$, not $4x$.

The roots are $\sqrt{16x^2} = 4x$ and $\sqrt{1} = 1$.

$16x^2 - 1 = (4x)^2 - 1^2 = (4x + 1)(4x - 1)$

The two factors have opposite signs.

Read $25 - 4x^2$ as $5^2 - (2x)^2$, with roots $5$ and $2x$.

$25 - 4x^2 = (5 + 2x)(5 - 2x)$

Keeping the order rebuilds $25 - 4x^2$; the form $(2x + 5)(2x - 5)$ equals $4x^2 - 25$.

The roots are $x$ and $y$.

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

Two variables work the same way as one.

The roots are $\sqrt{4x^2} = 2x$ and $\sqrt{9y^2} = 3y$.

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

Each root keeps its coefficient with its variable.

Neither term is a perfect square, so pull out the common factor $2$ first, then factor the difference of squares.

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

Stopping at $2(x^2 - 4)$ is incomplete because $x^2 - 4$ still factors.

Pull out the common factor $3$ first, then factor what remains.

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

The form $3(x^2 - 9)$ is not fully factored.

The product $(5x + 3)(5x - 3)$ is a difference of squares, so the middle terms cancel.

$(5x + 3)(5x - 3) = 25x^2 - 9$

There is no middle term, so it is not $25x^2 - 30x + 9$.

The roots are $\sqrt{9x^2} = 3x$ and $\sqrt{100} = 10$.

$9x^2 - 100 = (3x)^2 - 10^2 = (3x + 10)(3x - 10)$

The root of $9x^2$ is $3x$, not $9x$.

A sum of squares does not factor over the real numbers. Here $x^2 + 9$ is a sum of two squares.

$x^2 + 9 \text{ does not factor over the reals}$

The others are all differences of squares.

The roots are $\sqrt{16a^2} = 4a$ and $\sqrt{25b^2} = 5b$.

$16a^2 - 25b^2 = (4a)^2 - (5b)^2 = (4a + 5b)(4a - 5b)$

Each root keeps its coefficient with its variable.