Factoring by Grouping: Additional Practice (Set 2)

Here $ac = 6 \cdot (-20) = -120$ and the sum is $7$, met by $15$ and $-8$. Split $7x = 15x - 8x$.

$6x^2 + 15x - 8x - 20 = 3x(2x + 5) - 4(2x + 5) = (2x + 5)(3x - 4)$

Expanding gives $6x^2 + 7x - 20$.

Here $ac = 6 \cdot 12 = 72$ and the sum is $17$, met by $8$ and $9$. Split $17x = 8x + 9x$.

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

Expanding gives $6x^2 + 17x + 12$.

Pull out the common factor $2$ first, then group the four terms inside.

$2(x^3 + 3x^2 - 4x - 12) = 2[x^2(x + 3) - 4(x + 3)] = 2(x + 3)(x^2 - 4)$

The difference of squares $x^2 - 4 = (x - 2)(x + 2)$ factors again, giving $2(x + 3)(x - 2)(x + 2)$. Dropping the $2$ leaves out a factor of the original polynomial.

The first pair gives $x^2(x - 4)$, and the second pair $3x - 12 = 3(x - 4)$ matches.

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

Both pairs share $x - 4$.

Here $ac = 6 \cdot (-10) = -60$ and the sum is $-11$, met by $-15$ and $4$. Split $-11x = -15x + 4x$.

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

Expanding gives $6x^2 - 11x - 10$.

Put the terms in descending order: $3x^3 + 9x^2 + 2x + 6$. Then group.

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

Both pairs share $x + 3$.

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

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

The leftover $x^2 - 4 = (x - 2)(x + 2)$ repeats the factor $x - 2$, making it a square.

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

$4x^2(x + 1) - 9(x + 1) = (x + 1)(4x^2 - 9) = (x + 1)(2x - 3)(2x + 3)$

The binomial from grouping is $x + 1$, and $4x^2 - 9$, not $4x^2 + 9$, is what remains.

Here $ac = 10 \cdot 3 = 30$ and the sum is $11$, met by $6$ and $5$. Split $11x = 6x + 5x$.

$10x^2 + 6x + 5x + 3 = 2x(5x + 3) + 1(5x + 3) = (5x + 3)(2x + 1)$

Expanding gives $10x^2 + 11x + 3$.

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

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

Both pairs share $2x - 5$.

Here $ac = 20 \cdot 6 = 120$ and the sum is $-23$, met by $-15$ and $-8$. Split $-23x = -15x - 8x$.

$20x^2 - 15x - 8x + 6 = 5x(4x - 3) - 2(4x - 3) = (4x - 3)(5x - 2)$

Expanding gives $20x^2 - 23x + 6$.

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

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

The binomial from grouping is $3x + 2$, and $x^2 - 4$, not $x^2 + 4$, is what remains.