Factoring by Grouping: Additional Practice (Set 1)

Factor $-5$ from the second pair so its binomial matches $x - 3$.

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

Factoring $+5$ would give $5(3 - x)$, the wrong binomial.

Pull out the common factor $5$ first, then group the four terms inside.

$5(x^3 + 2x^2 - x - 2) = 5[x^2(x + 2) - 1(x + 2)] = 5(x + 2)(x^2 - 1)$

The difference of squares $x^2 - 1 = (x - 1)(x + 1)$ factors again, giving $5(x + 2)(x - 1)(x + 1)$. The factor is $x^2 - 1$, not $x^2 + 1$, and the $5$ must stay.

Here $ac = 9 \cdot 2 = 18$ and the sum is $9$, met by $6$ and $3$. Split $9x = 6x + 3x$.

$9x^2 + 6x + 3x + 2 = 3x(3x + 2) + 1(3x + 2) = (3x + 2)(3x + 1)$

Expanding gives $9x^2 + 9x + 2$.

Test each by grouping. The three that work leave matching binomials, for example $x^3 + 3x^2 + 5x + 15 = (x + 3)(x^2 + 5)$.

$x^3 + 2x^2 + 5x + 15 = x^2(x + 2) + 5(x + 3)$

Here the leftover binomials $x + 2$ and $x + 3$ differ, and no reordering fixes it, so this one does not group.

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

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

The binomial from grouping is $2x + 3$, and $x^2 - 1$, not $x^2 + 1$, is what remains.

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

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

The equal factors make this a perfect square.

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

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

The binomial from grouping is $x + 5$, and $x^2 - 4$, not $x^2 + 4$, is what remains.

Here $ac = 6 \cdot (-12) = -72$ and the sum is $-1$, met by $-9$ and $8$. Split $-x = -9x + 8x$.

$6x^2 - 9x + 8x - 12 = 3x(2x - 3) + 4(2x - 3) = (2x - 3)(3x + 4)$

Expanding gives $6x^2 - x - 12$.

Here $ac = 10 \cdot 2 = 20$ and the sum is $9$, met by $5$ and $4$. Split $9x = 5x + 4x$.

$10x^2 + 5x + 4x + 2 = 5x(2x + 1) + 2(2x + 1) = (2x + 1)(5x + 2)$

Expanding gives $10x^2 + 9x + 2$.

The two pieces must multiply to $ac = 12 \cdot 6 = 72$ and add to $-17$, met by $-8$ and $-9$.

$12x^2 - 8x - 9x + 6 = 4x(3x - 2) - 3(3x - 2) = (3x - 2)(4x - 3)$

The other splits add to $-17$ but multiply to $66$, $60$, or $42$, not $72$.

The first pair gives $x^2(x - 6)$, and the second pair $2x - 12 = 2(x - 6)$ matches.

$(x^3 - 6x^2) + (2x - 12) = x^2(x - 6) + 2(x - 6) = (x - 6)(x^2 + 2)$

Both pairs share $x - 6$.

Here $ac = 15 \cdot 8 = 120$ and the sum is $-22$, met by $-10$ and $-12$. Split $-22x = -10x - 12x$.

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

Expanding gives $15x^2 - 22x + 8$.