Difference of Squares: Additional Practice (Set 1)

The roots are $\sqrt{121x^2} = 11x$ and $\sqrt{49} = 7$.

$121x^2 - 49 = (11x)^2 - 7^2 = (11x + 7)(11x - 7)$

The root of $121x^2$ is $11x$, not $121x$.

Factor and use the zero-product property.

$x^2 - 64 = (x + 8)(x - 8) = 0 \implies x = 8 \text{ or } x = -8$

Both roots are solutions.

A sum of squares does not factor over the real numbers. Here $x^2 + 1$ is a sum of two squares.

$x^2 + 1 \text{ does not factor over the reals}$

The others are differences of squares.

Pull out the common factor $2$ first, then factor completely. The factor $x^2 - 4$ factors again.

$2x^4 - 32 = 2(x^4 - 16) = 2(x^2 + 4)(x + 2)(x - 2)$

The sum of squares $x^2 + 4$ stops there.

Write each factor as a distance from $100$: $105 = 100 + 5$ and $95 = 100 - 5$.

$105 \times 95 = (100 + 5)(100 - 5) = 100^2 - 5^2 = 10000 - 25 = 9975$

The product is a square minus $25$.

Read $144 - 25x^2$ as $12^2 - (5x)^2$, with roots $12$ and $5x$.

$144 - 25x^2 = (12 + 5x)(12 - 5x)$

Keeping the order rebuilds $144 - 25x^2$; the form $(5x + 12)(5x - 12)$ equals $25x^2 - 144$.

Pull out the common factor $7$ first, then factor the difference of squares.

$7x^2 - 28 = 7(x^2 - 4) = 7(x + 2)(x - 2)$

Stopping at $7(x^2 - 4)$ is incomplete.

Factor $9x^2 - 16$ as a difference of squares, then use the zero-product property.

$9x^2 - 16 = (3x + 4)(3x - 4) = 0 \implies x = \tfrac{4}{3} \text{ or } x = -\tfrac{4}{3}$

Each factor set to zero gives one solution.

Read $x^4 - 625$ as $(x^2)^2 - 25^2$. The factor $x^2 - 25$ factors again.

$x^4 - 625 = (x^2 + 25)(x^2 - 25) = (x^2 + 25)(x + 5)(x - 5)$

The sum of squares $x^2 + 25$ stops there.

The roots are $\sqrt{64x^2} = 8x$ and $\sqrt{81} = 9$.

$64x^2 - 81 = (8x)^2 - 9^2 = (8x + 9)(8x - 9)$

The root of $64x^2$ is $8x$, not $64x$.

The roots are $\sqrt{16a^2} = 4a$ and $\sqrt{9b^2} = 3b$.

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

Each root keeps its coefficient with its variable.

Pull out the common factor $5$ first, then factor completely. The factor $x^2 - 1$ factors again.

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

The sum of squares $x^2 + 1$ stops there.