Difference of Squares: Practice

Take the square root of each term: $\sqrt{x^2} = x$ and $\sqrt{9} = 3$. A difference of squares factors as the sum of the roots times their difference.

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

The two factors have opposite signs, so $(x - 3)^2$ is not correct.

The roots are $x$ and $\sqrt{1} = 1$.

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

This is the difference of squares with $b = 1$.

The roots are $x$ and $\sqrt{49} = 7$.

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

The two factors share the roots $x$ and $7$, once added and once subtracted.

Both $x^2$ and $4$ are perfect squares, but they are added, not subtracted. A sum of squares does not factor over the real numbers.

$x^2 + 4 \text{ does not factor over the reals}$

Only a difference of squares factors, so none of the product forms are correct.

A difference of squares needs both terms to be perfect squares. Here $7$ is not a perfect square, since $\sqrt{7}$ is not a whole number.

$x^2 - 7 \text{ is not a difference of squares}$

The others use $9$, $16$, and $81$, which are all perfect squares.

The roots are $x$ and $\sqrt{81} = 9$.

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

The factors have opposite signs.

Read $4 - x^2$ as $2^2 - x^2$, with roots $2$ and $x$. Keep the order so the product rebuilds $4 - x^2$.

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

The form $(x + 2)(x - 2)$ equals $x^2 - 4$, which is the opposite.

The roots are $x$ and $\sqrt{144} = 12$.

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

The factorization uses opposite signs.

Read $9 - x^2$ as $3^2 - x^2$, with roots $3$ and $x$, keeping the order.

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

The form $(x + 3)(x - 3)$ equals $x^2 - 9$, the opposite.

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

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

The factors have opposite signs.

The product $(x + 6)(x - 6)$ is a difference of squares, so the middle terms cancel.

$(x + 6)(x - 6) = x^2 - 36$

There is no middle term, so it is not $x^2 - 12x + 36$.

The roots are $x$ and $\sqrt{121} = 11$.

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

The factorization uses opposite signs.