Factoring by Grouping

Split the polynomial into the first two terms and the last two terms, then factor each pair on its own. The first pair shares $x^2$, and the second pair shares $3$:

Both pieces now carry the same binomial $x + 2$. By the rule just proved, factor it out:

Check by expanding: $(x + 2)(x^2 + 3) = x^3 + 3x + 2x^2 + 6 = x^3 + 2x^2 + 3x + 6$, the original polynomial. The remaining factor $x^2 + 3$ is a sum that stays positive for every real $x$, so it does not factor over the real numbers. That makes $(x + 2)(x^2 + 3)$ the complete factorization.

The pairing is not the only route. Grouping the first and third terms and the second and fourth terms works just as well, since $x^3 + 3x = x(x^2 + 3)$ and $2x^2 + 6 = 2(x^2 + 3)$ share the same binomial:

Any pairing that leaves matching binomials leads to the same answer.

Group the first two and last two terms. The first pair gives $x^2(x - 2)$ with no trouble:

Now look hard at the second pair, $-3x + 6$. Its terms share $3$, but factoring out $+3$ gives $3(-x + 2) = 3(2 - x)$, whose binomial $2 - x$ does not match the $x - 2$ from the first pair. Factor out $-3$ instead, so the binomial comes out as $x - 2$:

With both pairs carrying $x - 2$, the shared binomial factors out:

Check: $(x - 2)(x^2 - 3) = x^3 - 3x - 2x^2 + 6 = x^3 - 2x^2 - 3x + 6$. The factor $x^2 - 3$ has no factorization with integer coefficients, because a difference of squares needs a perfect square and $3$ is not one. So this is as far as grouping takes it. The rule to remember: when the second pair starts with a negative, factor a negative out so its binomial matches the first pair.

Take $x^3 + 6 + 2x^2 + 3x$, with its terms out of the usual order. Pairing them as they stand gives

and this stalls: $x^3 + 6$ has no common factor beyond $1$, so no shared binomial can appear. Rewrite the polynomial in descending order first:

Now the neighboring pairs cooperate, exactly as in Worked Example 1:

A grouping that stalls is often just a signal to reorder the terms, not a dead end.

Here $a = 2$, $b = 7$, and $c = 3$, so the product to aim for is $a \cdot c = 2 \cdot 3 = 6$. Find two numbers with product $6$ and sum $7$, namely $6$ and $1$. Split the middle term $7x$ into $6x + 1x$:

Now group the four terms and factor each pair:

Keep the $1$ in front of the second $x + 3$: it is the coefficient that becomes the $2x + 1$. Check by expanding: $(x + 3)(2x + 1) = 2x^2 + x + 6x + 3 = 2x^2 + 7x + 3$. This is the same factorization reverse FOIL produced last chapter, reached here without trying a single arrangement.

Now $a = 6$, $b = 7$, and $c = -3$, so $a \cdot c = 6 \cdot (-3) = -18$. Two numbers with product $-18$ and sum $7$ must have opposite signs, and $9$ and $-2$ fit, since $9 + (-2) = 7$ and $9 \cdot (-2) = -18$. Split $7x$ into $9x - 2x$:

Group, and mind the sign in the second pair. Its leading term $-2x$ is negative, so factor out $-1$:

Check: $(2x + 3)(3x - 1) = 6x^2 - 2x + 9x - 3 = 6x^2 + 7x - 3$. The order of the split does not matter; writing $6x^2 - 2x + 9x - 3$ instead groups to $2x(3x - 1) + 3(3x - 1) = (3x - 1)(2x + 3)$, the same factors. In the previous chapter this trinomial was factored by reverse-FOIL trial, and grouping reaches the identical answer straight down the middle.

Every coefficient is even, so factor out $2$ before doing anything else:

Now group inside the parentheses. The first pair shares $2x^2$ and the second shares $3$:

Carry the $2$ back out front for the complete factorization:

Skipping the common factor and grouping the original directly gives $4x^2(x + 2) + 6(x + 2) = (x + 2)(4x^2 + 6)$. The result $(x + 2)(4x^2 + 6)$ is a correct factorization but not a complete one, since $4x^2 + 6$ still hides a factor of $2$. Pulling the common factor out first, or at the very end, makes sure nothing is left behind.