Factoring by Grouping: Practice

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

$(4x^3 + 6x^2) + (10x + 15) = 2x^2(2x + 3) + 5(2x + 3) = (2x + 3)(2x^2 + 5)$

Both pairs share $2x + 3$.

Here $ac = 12 \cdot 12 = 144$ and the sum is $25$, met by $16$ and $9$. Split $25x = 16x + 9x$.

$12x^2 + 16x + 9x + 12 = 4x(3x + 4) + 3(3x + 4) = (3x + 4)(4x + 3)$

The other choices hide a common factor or give the wrong middle term.

Test each by grouping. The three that work leave matching binomials, for example $x^3 + 2x^2 + 3x + 6 = (x + 2)(x^2 + 3)$.

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

Here the leftover binomials $x + 1$ and $x + 3$ differ, and no reordering fixes it, so this one does not group.

Pull out the common factor $3$ first, then group the four terms inside.

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

The difference of squares $x^2 - 4 = (x - 2)(x + 2)$ factors again, giving $3(x - 1)(x - 2)(x + 2)$. The factor is $x^2 - 4$, not $x^2 + 4$, and the $3$ must stay.

Here $ac = 8 \cdot (-15) = -120$ and the sum is $2$, met by $12$ and $-10$. Split $2x = 12x - 10x$.

$8x^2 + 12x - 10x - 15 = 4x(2x + 3) - 5(2x + 3) = (2x + 3)(4x - 5)$

Expanding gives $8x^2 + 2x - 15$.

Here $ac = 9 \cdot 4 = 36$ and the sum is $-12$, met by $-6$ and $-6$. Split $-12x = -6x - 6x$.

$9x^2 - 6x - 6x + 4 = 3x(3x - 2) - 2(3x - 2) = (3x - 2)(3x - 2) = (3x - 2)^2$

The equal factors make this a perfect square.

Put the terms in descending order: $3x^3 - 6x^2 - 8x + 16$. Then group.

$(3x^3 - 6x^2) + (-8x + 16) = 3x^2(x - 2) - 8(x - 2) = (x - 2)(3x^2 - 8)$

The second pair is $-8x + 16 = -8(x - 2)$.

Here $ac = 15 \cdot (-6) = -90$ and the sum is $1$, met by $10$ and $-9$. Split $x = 10x - 9x$.

$15x^2 + 10x - 9x - 6 = 5x(3x + 2) - 3(3x + 2) = (3x + 2)(5x - 3)$

Expanding gives $15x^2 + x - 6$.

Group first, then factor the difference of squares that appears.

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

The binomial from grouping is $x + 4$, not $x - 4$, and $x^2 - 9 = (x - 3)(x + 3)$.

The two pieces must multiply to $ac = 6 \cdot 6 = 36$ and add to $13$, met by $4$ and $9$.

$6x^2 + 4x + 9x + 6 = 2x(3x + 2) + 3(3x + 2) = (3x + 2)(2x + 3)$

The other splits add to $13$ but multiply to $42$, $30$, or $40$, not $36$.

Group first, factoring $-1$ from the second pair, then factor the difference of squares.

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

The binomial from grouping is $x - 2$, and $4x^2 - 1$, not $4x^2 + 1$, is what remains.

Group first, then factor the difference of squares that appears.

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

The binomial from grouping is $2x + 1$, and $x^2 - 9$, not $x^2 + 9$, is what remains.