Factoring by Grouping: Practice

Both terms share the binomial $x + 5$. Factor it out; the leftover pieces $x^2$ and $2$ form the other factor.

$x^2(x + 5) + 2(x + 5) = (x + 5)(x^2 + 2)$

The $+2$ stays a plus, so $(x + 5)(x^2 - 2)$ is a sign error.

Group the first two and last two terms, then factor each pair.

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

Both pairs leave $x + 4$, which factors out.

Factor $x^2$ from the first pair and $5$ from the second pair.

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

The shared binomial $x + 2$ comes out front.

Both terms are written as something times $x - 7$, so that binomial is the common factor.

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

The leftover $x^2 + 4$ is the other factor, not the common one.

The greatest common factor of $5x$ and $15$ is $5$. Divide each term by $5$.

$5x + 15 = 5(x + 3)$

Expanding $5(x + 3) = 5x + 15$ confirms it.

Factor $2x$ from the first pair and $3$ from the second pair.

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

Both pairs share $x + 4$.

Both terms carry $x + 2$, so factor it out; the leftover $x^2$ and $4$ form the second factor.

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

The second factor keeps the $x^2$, so $(x + 2)(x + 4)$ is wrong.

Factor $x^2$ from the first pair and $4$ from the second pair.

$(x^3 + 5x^2) + (4x + 20) = x^2(x + 5) + 4(x + 5) = (x + 5)(x^2 + 4)$

The shared binomial is $x + 5$.

Factor $x^2$ from the first pair and $5$ from the second pair.

$(x^3 + 9x^2) + (5x + 45) = x^2(x + 9) + 5(x + 9) = (x + 9)(x^2 + 5)$

The shared binomial is $x + 9$.

The split needs two numbers with product $ac = 8$ and sum $9$.

$8 \cdot 1 = 8, \qquad 8 + 1 = 9$

The other pairs sum to $9$ but multiply to $18$, $20$, or $14$, not $8$.

Factor $2x$ from the first pair and $1$ from the second pair.

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

Keep the $1$; it becomes the constant in $2x + 1$.

Factor $x^2$ from the first pair and $3$ from the second pair.

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

The common binomial is $x + 8$.