Factoring by Grouping: Free Response

Group the first two terms and the last two terms.

$x^3 + 6x^2 + 5x + 30 = (x^3 + 6x^2) + (5x + 30).$

The first pair shares $x^2$: $x^3+6x^2 = x^2(x+6)$. The second pair shares $5$: $5x+30 = 5(x+6)$.

Both pairs from part A carry the same binomial, so it factors out in front.

$x^2(x+6) + 5(x+6) = (x+6)(x^2+5).$

Expanding checks the answer: $(x+6)(x^2+5) = x^3+5x+6x^2+30 = x^3+6x^2+5x+30$, the original polynomial.

Factoring each pair only sets up the method; grouping finishes only when the two leftover pieces are the identical binomial, since that is exactly what the distributive property needs:

$ka + kb = k(a+b).$

If the first pairing leaves two different binomials, that particular pairing has failed, but the polynomial has not necessarily failed: reordering the four terms, most often into descending order by degree, or trying a different pairing, can still produce a match. Only after checking the reasonable pairings is it fair to conclude the polynomial does not group at all.

Here $a=2$, $b=11$, $c=12$, so the split needs two numbers with product $ac=2\cdot12=24$ and sum $11$. Those numbers are $8$ and $3$, since $8\cdot3=24$ and $8+3=11$. Split the middle term:

$2x^2+11x+12 = 2x^2+8x+3x+12.$

Group and factor each pair:

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

Set each factor from part A equal to $0$.

$x+4=0 \quad\Rightarrow\quad x=-4, \qquad 2x+3=0 \quad\Rightarrow\quad x=-\frac{3}{2}.$

Both values solve the original equation; neither factor is dropped.

Reversing the order gives

$2x^2+3x+8x+12.$

Grouping this the same way: $(2x^2+3x)+(8x+12) = x(2x+3)+4(2x+3) = (2x+3)(x+4)$, the same two factors as part A, just written in the other order. The middle term $11x$ is the sum $8x+3x$, and addition gives the same total no matter which piece is written first, so the four-term line itself is unchanged in value, only in the order its two middle terms are listed. Grouping a rearranged sum of the same four terms cannot change what it factors to.

Group the first two and last two terms.

$x^3-4x^2-6x+24 = (x^3-4x^2)+(-6x+24).$

The first pair gives $x^2(x-4)$. The second pair, $-6x+24$, leads with a negative term: factoring out $+6$ would give $6(4-x)$, which does not match $x-4$, so factor out $-6$ instead.

$-6x+24 = -6(x-4).$

Both pairs now carry $x-4$:

$x^2(x-4)-6(x-4) = (x-4)(x^2-6).$

Expanding checks it: $(x-4)(x^2-6)=x^3-6x-4x^2+24=x^3-4x^2-6x+24$.

Try the terms in their given order first.

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

The leftover binomials, $x+3$ and $x+4$, do not match. Try pairing the first and third terms with the second and fourth:

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

The second group has no common factor beyond $1$, and it is not of the matching form either, so this pairing fails too. The remaining pairing, the first and fourth terms with the second and third, fares no better: $x^3+20$ shares no common factor with $3x^2+5x$ that could line up as a matching binomial. With all three ways of splitting four terms into two pairs tried, none leaves a shared binomial, so this polynomial does not factor by grouping.

Every pairing produces two leftover pieces once each pair's greatest common factor is removed. The test that decides success has nothing to do with the size of the coefficients or how many terms cancel; it is simply whether those two leftover pieces are the same binomial.

$k_1(x+m) + k_2(x+m) = (x+m)(k_1+k_2)$

When the pieces match, the identical binomial factors out by the distributive property. When they do not match, as in every pairing tried, there is nothing to pull out, and no amount of further arithmetic manufactures a shared factor. Checking every reasonable pairing this way is the only honest route to concluding that a polynomial fails to group, rather than assuming it after one attempt.

Test each step for what it actually establishes.

Step 1 is the trinomial itself: nothing to check. Step 2 claims $-9$ and $5$ are the numbers the AC method needs, backed only by $-9+5=-4$. The AC method needs a pair satisfying two conditions at once, sum $b=-4$ and product $ac=-60$, and Step 2 never checks the second one.

$(-9)(5) = -45 \neq -60.$

So Step 2 is the first step that is not fully justified, well before Step 5's false conclusion. Steps 3 and 4 carry out correct algebra on the wrong pair from Step 2, which is why they do not themselves contain an error, only an unhelpful result: the mismatched binomials $4x-9$ and $x-3$ are the visible symptom of Step 2's mistake, not a new mistake of their own. A pair that actually satisfies both conditions is $6$ and $-10$, since $6\cdot(-10) = -60$ and $6+(-10)=-4$.

Split the middle term with the pair from part A.

$4x^2-4x-15 = 4x^2+6x-10x-15.$

Group and factor each pair, minding the sign on the second one:

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

Expanding checks it: $(2x+3)(2x-5) = 4x^2-10x+6x-15 = 4x^2-4x-15$, the original trinomial.

Grouping only completes when the two leftover pieces are the identical binomial. A pair must satisfy both conditions at once:

$m+n=b, \qquad mn=ac.$

If the pieces after grouping are not the identical binomial, that is not a sign to keep pushing forward with what is at hand; it is evidence that the split itself was wrong, specifically, that the numbers used, while adding to $b$, must not actually multiply to $ac$, since a pair meeting both conditions is exactly what guarantees the four terms will group. The correct response is to go back to the search for numbers with product $ac$ and sum $b$, check both conditions this time, and split again with the corrected pair, rather than combining mismatched pieces into an invalid product.

Every term shares a factor of $3x$.

$6x^3-3x^2-45x = 3x(2x^2-x-15).$

Inside the parentheses, $a=2$, $b=-1$, $c=-15$, so $ac = 2\cdot(-15) = -30$. Two numbers with product $-30$ and sum $-1$ are $5$ and $-6$, since $5\cdot(-6)=-30$ and $5+(-6)=-1$.

Split the middle term with the pair from part A.

$2x^2-x-15 = 2x^2+5x-6x-15.$

Group, factoring a negative out of the second pair so its binomial matches the first:

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

Carrying the $3x$ from part A back through gives the complete factorization:

$6x^3-3x^2-45x = 3x(2x+5)(x-3).$

Set every factor of $3x(2x+5)(x-3)$ equal to $0$ in turn, including the leading $3x$, not only the two binomials.

$3x=0 \ \Rightarrow\ x=0, \qquad 2x+5=0 \ \Rightarrow\ x=-\frac{5}{2}, \qquad x-3=0 \ \Rightarrow\ x=3.$

Three different factors give three different solutions.