Factoring Quadratics: Free Response

The product $c = -24$ is negative, so the two numbers have opposite signs, and the sum $b = -2$ is negative, so the larger one (by size) is negative. List the pairs of integers that multiply to $-24$ and check their sums.

$(-6)(4) = -24, \quad -6 + 4 = -2.$

That pair fits, so

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

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

Every coefficient is divisible by $3$, so pull that factor out first.

$3x^2 + 27x + 54 = 3\left(x^2 + 9x + 18\right).$

Inside the parentheses, $c = 18$ is positive and $b = 9$ is positive, so both numbers are positive. Two positive numbers that multiply to $18$ and add to $9$ are $3$ and $6$:

$x^2 + 9x + 18 = (x + 3)(x + 6).$

So the complete factorization is $3(x + 3)(x + 6)$; the $3$ pulled out front is part of the answer, not a step you can drop.

The sign of $c$ decides whether the two numbers share a sign, because $c$ is their product.

$c = pq.$

A product of two numbers is negative only when the numbers have opposite signs, and positive only when they share a sign (both positive or both negative). In part A, $c = -24 < 0$ forced opposite signs. Inside part B's parentheses, $c = 18 > 0$ forced a shared sign. Once a shared sign is guaranteed, the sign of $b = p + q$ decides which one: two numbers with the same sign add to a sum of that same sign, so $b = 9 > 0$ meant the shared sign was positive.

The product $c = -21$ is negative, so the two numbers have opposite signs, and the sum $b = 4$ is positive, so the larger one is positive. Test integer pairs multiplying to $-21$:

$(7)(-3) = -21, \quad 7 + (-3) = 4.$

That pair fits, so

$x^2 + 4x - 21 = (x + 7)(x - 3) = 0.$

By the zero-product property, $x + 7 = 0$ or $x - 3 = 0$, so $x = -7$ or $x = 3$. Each root is the OPPOSITE sign of the number written in its factor.

The factorization from part A names two numbers, $p$ and $q$, written inside the factors. Copying those two numbers directly, signs unchanged, is NOT the same step as solving the equation.

The zero-product property sets each FACTOR equal to $0$, not each number inside it:

$x + 7 = 0 \quad \text{or} \quad x - 3 = 0,$

which gives $x = -7$ or $x = 3$, the opposite of each number written in its factor. Copying the factor numbers directly instead would report $x = 7$ and $x = -3$, a pair with both signs flipped from the true roots. That pair does not satisfy the equation; only $x = -7$ and $x = 3$ do.

The factorization $(x + p)(x + q) = 0$ names two numbers, $p$ and $q$, but the equation is solved by asking what value of $x$ makes each FACTOR equal to $0$, not by reading off $p$ and $q$ themselves.

$x + p = 0 \quad \Longrightarrow \quad x = -p,$

and the same reasoning gives $x = -q$ from the other factor. The sign flip is not a memorized rule bolted onto the method; it falls straight out of solving $x + p = 0$ for $x$, and it happens for every value of $p$, not just the ones that appear in any one example.

The product $c = 10$ is positive, so any working pair would share a sign, matching the sign of $b = 6$: both positive. List the positive pairs that multiply to $10$ and check their sums.

$1 \times 10 = 10 \left(\text{sum } 11\right), \quad 2 \times 5 = 10 \left(\text{sum } 7\right).$

Those are the only two ways to write $10$ as a product of positive integers, and neither sums to $6$. Since $c > 0$ rules out a mixed-sign pair entirely, no integer pair has sum $6$ and product $10$, so $x^2 + 6x + 10$ does not factor over the integers.

Here $c = 9$ is positive and $b = 6$ is positive, so both numbers are positive. Two positive numbers with product $9$ and sum $6$ are $3$ and $3$:

$3 \times 3 = 9, \quad 3 + 3 = 6.$

So $x^2 + 6x + 9 = (x + 3)(x + 3)$. Nothing in the method requires the two numbers to be different from each other.

Set each factor from part B equal to $0$:

$x + 3 = 0 \quad \text{or} \quad x + 3 = 0,$

and both give the same value, $x = -3$. The zero-product property does not promise two DIFFERENT solutions; it only promises that the product is zero when at least one factor is. Here both factors happen to be the same, so the method still runs correctly, it simply lands on one value instead of two. Saying the equation has a repeated root, one solution counted twice, is accurate; saying it has two different solutions is not.

Every coefficient is even, so pull the common factor $2$ out first.

$2x^2 + 4x - 96 = 2\left(x^2 + 2x - 48\right).$

Inside, $c = -48$ is negative, so the two numbers have opposite signs, and $b = 2$ is positive, so the larger one is positive. Test pairs multiplying to $-48$:

$(8)(-6) = -48, \quad 8 + (-6) = 2.$

That pair fits, so $x^2 + 2x - 48 = (x + 8)(x - 6)$, and the complete factorization is $2(x + 8)(x - 6)$.

The two counts are $x + 8$ and $x - 6$. For a real stack, both must be positive:

$x + 8 > 0 \quad \text{and} \quad x - 6 > 0.$

The first holds for every $x > -8$, no real restriction once $x$ is a positive order size. The second is the binding condition: it requires $x > 6$. The smallest whole number greater than $6$ is $x = 7$; at $x = 6$ the second count would be exactly $0$, so the stack would hold no tiles at all.

Substitute $x = 10$ into each linear factor from part A:

$x + 8 = 18, \qquad x - 6 = 4.$

The total from the factored form is

$2 \times 18 \times 4 = 144 \text{ tiles}.$

Checking directly from the original expression,

$2(10)^2 + 4(10) - 96 = 200 + 40 - 96 = 144,$

the same value. The two routes agree because factoring never changed the expression, only its form.

The claim is about every integer pair with a negative product, so the argument has to rule out every same-sign case, not check just one example.

$pq = c < 0.$

Since $pq \neq 0$, neither $p$ nor $q$ is $0$. That leaves the two same-sign cases to rule out, plus the mixed case.

If $p > 0$ and $q > 0$, a positive times a positive is positive, so $pq > 0$, contradicting $pq < 0$.

If $p < 0$ and $q < 0$, a negative times a negative is also positive, so again $pq > 0$, the same contradiction.

Both same-sign cases are impossible, so the only possibility left is that one of $p, q$ is positive and the other is negative: opposite signs.

By part A, $p$ and $q$ have opposite signs. Let the one with the larger absolute value have size $m$ and the other have size $n$, with $m > n > 0$ (if their absolute values were equal the sum would be $0$, which has no sign and leaves the claim with nothing to assert, so that case is set aside rather than ruled out: $p = 3$, $q = -3$ is a genuine pair of that kind).

Case 1: the larger one is positive. Then $p = m$ and $q = -n$, so

$b = p + q = m - n.$

Since $m > n$, this is positive, matching the sign of the larger number.

Case 2: the larger one is negative. Then $p = -m$ and $q = n$, so

$b = p + q = -m + n = -(m - n),$

which is negative since $m > n$, again matching the sign of the larger number.

Both cases give $b$ the same sign as whichever of $p, q$ has the larger absolute value, which is the claim.

From the coefficients, $b = -5$ and $c = -36$. Since $c < 0$, part A says the two numbers have opposite signs. Since $b < 0$, part B says the one with the larger absolute value is negative. The prediction, made before any search, is: opposite signs, with the bigger one negative.

Now search for the actual pair: integers with product $-36$ and sum $-5$.

$(4)(-9) = -36, \quad 4 + (-9) = -5.$

The pair is $4$ and $-9$: one positive number and one negative number, opposite signs as predicted, and the larger in absolute value, $9$, carries a minus sign, also as predicted. The two general facts, proved once, correctly called the outcome for a trinomial neither part A nor part B ever mentioned by name.