Factoring Harder Quadratics: Free Response

The middle term should split into two numbers with sum $11$ and product $a\cdot c = 2 \times 12 = 24$.

$8 + 3 = 11 \qquad \text{and} \qquad 8 \times 3 = 24.$

So the two middle strips are $8x$ and $3x$. Pairing $x$ with $3$ gives a strip of $3x$, and pairing $2x$ with $4$ gives a strip of $8x$, which suggests $(2x+3)(x+4)$. Expand to confirm:

$(2x+3)(x+4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12.$

Evaluate each factor from part A at $x=3$: $2(3)+3 = 9$ feet and $3+4 = 7$ feet.

$9 \times 7 = 63.$

Evaluate the area formula directly at $x=3$: $2(3)^2+11(3)+12 = 18+33+12=63$. The two match, confirming the factorization.

Solve the inequality:

$x + 4 \ge 8$

gives $x \ge 4$ by subtracting $4$ from both sides. The smallest integer satisfying this is $x=4$, which gives a side length of exactly $8$ feet. The next smaller integer, $x=3$, gives a side length of $3+4=7$ feet, which is below the required $8$ feet, so it fails the requirement.

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

Compare term by term with the original $3x^2 - 4x - 4$: the leading term $3x^2$ matches, and the constant $-4$ matches, but the middle term is $+4x$ where it should be $-4x$, a difference of $8x$. So the classmate's factorization is wrong, even though it uses the correct numbers.

The two cross products need to be built from the factor pair $3,1$ of $a$ and a factor pair of $c=-4$ so that they come out to $2x$ and $-6x$. Pairing $3x$ with $-2$ gives a strip of $-6x$, and pairing $x$ with $2$ gives a strip of $2x$, the reverse of the classmate's assignment:

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

The middle term now comes out to $-4x$, matching the original, so this is the correct factorization.

From $(3x+2)(x-2)=0$:

$3x+2=0 \qquad \text{or} \qquad x-2=0,$

giving $x = -\tfrac{2}{3}$ or $x = 2$. Check the fraction: $3\left(-\tfrac{2}{3}\right)^2 - 4\left(-\tfrac{2}{3}\right) - 4 = 3 \cdot \tfrac{4}{9} + \tfrac{8}{3} - 4 = \tfrac{4}{3}+\tfrac{8}{3}-4 = 4-4=0$. Both roots hold.

The classmate's product $(3x-2)(x+2)$ is not a rewriting of $3x^2-4x-4$ at all: part A showed it expands to $3x^2+4x-4$, a different trinomial with a different middle coefficient. Setting a DIFFERENT expression to zero solves a DIFFERENT equation, whose roots need not have anything to do with the roots of $3x^2-4x-4=0$. Matching constant terms is not enough: the middle term has to match too, or the two quadratics are not the same equation at all.

The three coefficients $4$, $-2$, and $-30$ share a common factor of $2$:

$4x^2 - 2x - 30 = 2\left(2x^2 - x - 15\right).$

Now factor $2x^2 - x - 15$. The middle term should split into two numbers with sum $-1$ and product $a\cdot c = 2\times(-15) = -30$:

$5 + (-6) = -1 \qquad \text{and} \qquad 5\times(-6) = -30.$

Pairing $2x$ with $5$ and $x$ with $-3$ gives $(2x+5)(x-3)$. Expand to confirm:

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

Putting the factor of $2$ back gives the complete factorization:

$4x^2 - 2x - 30 = 2(2x+5)(x-3).$

$2(2x+5)(x-3)=0$. The constant factor $2$ is never zero, so the zero-product property applies to the two binomials:

$2x+5=0 \qquad \text{or} \qquad x-3=0,$

giving $x = -\tfrac{5}{2}$ or $x=3$. Check the fraction: $4\left(-\tfrac{5}{2}\right)^2 - 2\left(-\tfrac{5}{2}\right) - 30 = 4\cdot\tfrac{25}{4}+5-30=25+5-30=0$. Both roots hold.

Solving an equation $2(2x+5)(x-3)=0$ only asks which factor can be zero. The constant $2$ is never zero for any $x$, so it contributes nothing to the solution set: $x=-\tfrac{5}{2}$ or $x=3$ regardless of whether the $2$ is written down.

Factoring the EXPRESSION is a different demand: it asks for something that equals $4x^2-2x-30$ exactly, for every value of $x$, not just at the roots. Check what happens without the $2$:

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

which is HALF of $4x^2-2x-30$, not the same expression. So dropping the $2$ from the equation costs nothing, because the equation only cares about where the product is zero, while dropping it from the factorization produces a wrong expression, because the expression has to match everywhere, not just at its roots.

$3$, $2$, and $5$ have no common integer factor other than $1$, so nothing can be pulled out front.

Every integer pair with product $15$ is one of $1\ \&\ 15$, $3\ \&\ 5$, $-1\ \&\ -15$, or $-3\ \&\ -5$ (a positive product forces both numbers to share a sign). Their sums are

$1+15=16, \qquad 3+5=8, \qquad -1+(-15)=-16, \qquad -3+(-5)=-8.$

None of these is $2$, so no integer pair has product $15$ and sum $2$.

Part A showed that no integer pair has product $15$ and sum $2$: the four possible sums were

$16,\ 8,\ -16,\ -8,$

and none of them is $2$. Since the reverse-FOIL search depends entirely on finding such a pair, and none exists, $3x^2+2x+5$ does not factor over the integers.

That makes $3x^2+2x+5$ exactly the kind of case the claim in the stem says cannot happen: it has no common factor across its coefficients, yet it does not factor into integer binomials. One case where the claim's condition holds and its conclusion fails is enough to refute a claim stated for every such trinomial, so the claim is false as stated.

For $4x^2+4x+1$: $a\cdot c = 4\times1=4$ and $b=4$. Search for two numbers with product $4$ and sum $4$: $2$ and $2$ work ($2\times2=4$, $2+2=4$). Build the factors: with $p=q=2$ (a factor pair of $4$) and $r=s=1$ (a factor pair of $1$),

$(2x+1)(2x+1) = 4x^2 + 2x + 2x + 1 = 4x^2 + 4x + 1.$

This trinomial has no common factor across $4$, $4$, $1$ either, yet it DOES factor, into the same binomial written twice (a repeated root, $x=-\tfrac12$ counted twice).

Put the two cases side by side. $3x^2+2x+5$: no common factor, does NOT factor. $4x^2+4x+1$: no common factor, DOES factor. The presence or absence of a common factor decided nothing in either case; whether integers exist with the right sum and product is a separate question entirely. So having no common factor is neither a guarantee that a trinomial factors nor a guarantee that it does not: the claim in the stem, which asserted the first direction, is false.

$20$, $-12$, and $-8$ share a common factor of $4$:

$20x^2-12x-8 = 4\left(5x^2 - 3x - 2\right).$

Factor $5x^2-3x-2$: the middle term should split into two numbers with sum $-3$ and product $a\cdot c = 5\times(-2)=-10$:

$2 + (-5) = -3 \qquad \text{and} \qquad 2\times(-5)=-10.$

Pairing $5x$ with $2$ and $x$ with $-1$ gives $(5x+2)(x-1)$. Expand to confirm:

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

Putting the $4$ back gives the complete factorization $20x^2-12x-8=4(5x+2)(x-1)$. The constant $4$ is never zero, so

$5x+2=0 \qquad \text{or} \qquad x-1=0,$

giving $x=-\tfrac{2}{5}$ or $x=1$.

Two integers with product $-160$ and sum $-12$: since the product is large and negative, try a pair where one factor is much larger than the other. $8$ and $-20$ work: $8\times(-20)=-160$ and $8+(-20)=-12$.

Build a factor pair of $a=20$ and of $c=-8$ whose cross products are $8x$ and $-20x$. Using $p=5$, $q=4$ (a factor pair of $20$) with $r=2$, $s=-4$ (a factor pair of $-8$):

$(5x+2)(4x-4) = 20x^2 - 20x + 8x - 8 = 20x^2 - 12x - 8.$

The expansion matches, so $(5x+2)(4x-4)$ is a valid factorization of the original expression.

Part A's search worked with the small numbers $2$ and $-5$, from a product of only $-10$. Part B's search needed $8$ and $-20$, from a product of $-160$, sixteen times larger. Pulling the common factor out first keeps every number in the search smaller and easier to work with.

$(5x+2)(4x-4)$ does expand correctly back to the original expression, so it is not wrong: it really does multiply out to $20x^2-12x-8$. But look inside the second factor: $4x-4=4(x-1)$ still has a common factor of $4$ hiding in it. Pulling that $4$ out front,

$(5x+2)(4x-4) = (5x+2)\cdot 4(x-1) = 4(5x+2)(x-1),$

which is exactly the completely factored form from part A. So the two routes do not disagree: part B's result is simply an incompletely factored version of part A's, one more common-factor pull away from being finished. For SOLVING the equation neither version matters, since a factor of $4$ is never zero either way, but only part A's form is the complete factorization of the expression.