Solving Quadratics by Factoring: Free Response

The Zero Product Property is a statement about a product that equals ZERO and about no other number, so nothing can be done with this equation in the shape it arrives in. Expand the left side first:

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

Now move the $22$ across, so that one side really is zero:

$x^2 + 3x - 18 - 22 = 0, \qquad x^2 + 3x - 40 = 0.$

Factor what is left. A pair with product $-40$ and sum $3$ is $8$ and $-5$, so

$x^2 + 3x - 40 = (x + 8)(x - 5) = 0.$

The product of the two real numbers $x + 8$ and $x - 5$ is zero, so at least one of them is zero, which gives $x = -8$ or $x = 5$.

Test both in the printed equation rather than in the factored one you built, since a slip in the expanding would leave the built equation agreeing with itself:

$(5 - 3)(5 + 6) = 2 \cdot 11 = 22.$

$(-8 - 3)(-8 + 6) = (-11)(-2) = 22.$

Standard form is $y = x^2 + 3x - 40$, and the $y$-intercept is the value at $x = 0$, which is just the constant term:

$y(0) = -40.$

Factored form is $y = (x + 8)(x - 5)$, and its zeros are on display, $-8$ and $5$. Those are the $x$-intercepts, because a graph point $(x, y)$ sits on the $x$-axis exactly when $y = 0$.

The axis of symmetry is the midpoint of the two zeros, since the parabola reflects one crossing onto the other and the mirror must stand halfway between them:

$\frac{-8 + 5}{2} = -\frac{3}{2}.$

Standard form agrees, since $-\tfrac{b}{2a}$ with $a = 1$ and $b = 3$ is the same line. Neither form is the better one. Each answered a different question for nothing, which is the whole reason for changing form.

Test each candidate by substitution, with $k$ left standing, so that one calculation covers every value of $k$ at once.

First candidate, $x = k + 3$. Substituting into the left side gives $(k + 3 - 3)(k + 3 + 6)$, that is $k(k + 9)$, and the equation demands that this equal $k$:

$k(k + 9) = k \quad \Longrightarrow \quad k^2 + 8k = 0.$

$k(k + 8) = 0 \quad \Longrightarrow \quad k = 0 \ \text{ or } \ k = -8.$

Notice the move that was needed there: everything went to one side before anything was split, which is exactly the step the student skipped.

Second candidate, $x = k - 6$. Now $(k - 6 - 3)(k - 6 + 6)$ is $(k - 9)k$, and the equation demands

$(k - 9)k = k \quad \Longrightarrow \quad k^2 - 10k = 0.$

$k(k - 10) = 0 \quad \Longrightarrow \quad k = 0 \ \text{ or } \ k = 10.$

So the values are $k = 0$, $k = -8$ and $k = 10$.

What happens at each. At $k = 0$ both candidates are genuine, and that is the licensed case: the move happens to agree with the Zero Product Property. The other two are accidents. At $k = -8$ the equation becomes $x^2 + 3x - 10 = 0$, that is $(x + 5)(x - 2) = 0$, with roots $-5$ and $2$; the candidate $-5$ is a root, the candidate $-14$ is not, and the root $2$ never appears. At $k = 10$ the equation becomes $x^2 + 3x - 28 = 0$, that is $(x + 7)(x - 4) = 0$, with roots $4$ and $-7$; the candidate $4$ is a root, the candidate $13$ is not, and $-7$ never appears.

What this settles. It settles that the move is not wrong in the weak sense of never naming a solution. It settles nothing in its favour. A step is licensed when the reasoning forces its conclusion, and nothing here forces anything: for every $k$ except $0$ the move reports at least one number that is not a solution, and whenever the equation has any solutions at all it hides one of them, and it cannot tell you which is which. Only a right-hand side of $0$ turns the split into a deduction, because $0$ is the only number that cannot be written as a product of two nonzero reals.

Confirm the zero first, because the identity says nothing whatever about a number that is not one:

$f(-2) = 5(4) + 3(-2) - 14 = 20 - 6 - 14 = 0.$

Now read the identity with $a = 5$, $b = 3$ and $r = -2$, each with its own sign. The leftover factor is $ax + b + ar$:

$5x + 3 + 5(-2) = 5x - 7.$

So $f(x) = (x + 2)(5x - 7)$. Check by expanding:

$(x + 2)(5x - 7) = 5x^2 - 7x + 10x - 14 = 5x^2 + 3x - 14.$

No pair of numbers was hunted for anywhere in that. One known zero was enough to produce the whole factorization.

The Zero Product Property applies, because the product $(x + 2)(5x - 7)$ is set equal to zero:

$x + 2 = 0 \quad \text{or} \quad 5x - 7 = 0.$

The first gives $x = -2$ and the second gives $x = \tfrac{7}{5}$, so the solution set is $\left\{-2, \tfrac{7}{5}\right\}$. A fraction is a perfectly good root and is left as a fraction.

The axis of symmetry is the midpoint of the two zeros:

$\frac{-2 + \tfrac{7}{5}}{2} = \frac{-\tfrac{3}{5}}{2} = -\frac{3}{10}.$

Standard form gives the same line, since $-\tfrac{b}{2a}$ with $a = 5$ and $b = 3$ is $-\tfrac{3}{10}$. The two routes agree because the crossings are mirror images of each other, so the mirror stands halfway between them.

Set the second factor's value at $x = r$ equal to zero and solve the condition that results:

$ar + b + ar = 0 \quad \Longrightarrow \quad b = -2ar.$

$r = -\frac{b}{2a}.$

So the coincidence is not an accident of arithmetic. It happens exactly when the zero sits on the axis of symmetry, and the converse holds too: if $r$ is a zero and $r = -\tfrac{b}{2a}$, then $2ar + b = 0$, so the second factor does vanish there.

Substitute $b = -2ar$ back into the leftover factor:

$ax + b + ar = ax - 2ar + ar = a(x - r),$

so the identity collapses to $f(x) = a(x - r)^2$.

The solution set. Both factors now say the same thing, so $f(x) = 0$ has the one-element solution set $\{r\}$, even though the factor was collected twice. A set does not count multiplicity, so nothing about the set remembers the repetition.

The graph. The graph remembers instead. Take $a > 0$: a square is never negative, so $f$ lies on or above the $x$-axis and meets it only at $(r, 0)$, which is therefore the vertex. The parabola touches the axis there and turns back rather than crossing, so it is tangent to the axis, and the sign of $f$ does not change as $x$ passes $r$. For $a < 0$ everything is the same upside down.

On $g(x) = 9x^2 - 30x + 25$. Here $a = 9$, $b = -30$ and $c = 25$, so the prediction is $r = -\tfrac{b}{2a} = \tfrac{30}{18} = \tfrac{5}{3}$. Check that it is a zero:

$g\left(\tfrac{5}{3}\right) = 9 \cdot \tfrac{25}{9} - 30 \cdot \tfrac{5}{3} + 25 = 25 - 50 + 25 = 0.$

The leftover factor is $9x - 30 + 9\left(\tfrac{5}{3}\right) = 9x - 15$, so

$g(x) = \left(x - \tfrac{5}{3}\right)(9x - 15) = 9\left(x - \tfrac{5}{3}\right)^2 = (3x - 5)^2,$

the leading $9$ folded into the bracket. One zero, at $\tfrac{5}{3}$, and a parabola resting on the axis there.

Work out $(x - 3)(x - 6)$ for each member in turn, subtracting inside the system first and then multiplying inside it.

$x = 0: (7)(4) = 28 \to 8, \qquad x = 1: (8)(5) = 40 \to 0.$

$x = 2: (9)(6) = 54 \to 4, \qquad x = 3: (0)(7) = 0.$

$x = 4: (1)(8) = 8, \qquad x = 5: (2)(9) = 18 \to 8.$

$x = 6: (3)(0) = 0, \qquad x = 7: (4)(1) = 4.$

$x = 8: (5)(2) = 10 \to 0, \qquad x = 9: (6)(3) = 18 \to 8.$

Four members give $0$: $1$, $3$, $6$ and $8$.

Over the real numbers the same factored equation has exactly two solutions, the two numbers on display in its factors. Here those two are still solutions, and two more have appeared beside them.

Take $x = 8$. Its factors are $8 - 3 = 5$ and $8 - 6 = 2$, neither of them $0$, and yet

$5 \times 2 = 10 \to 0.$

The member $x = 1$ does the same job, since its factors are $8$ and $5$ and $8 \times 5 = 40 \to 0$. So this system contains nonzero members whose product is zero, which is exactly what the Zero Product Property forbids over the reals.

Now the proof. It ran: given $uv = 0$ and $u \neq 0$, multiply both sides by $\tfrac{1}{u}$, regroup, and read off $v = 0$. Every step of that survives here except the one that produced $\tfrac{1}{u}$ in the first place, namely that every nonzero number has a reciprocal. Test it on $5$:

$5 \times 0 = 0, \quad 5 \times 1 = 5, \quad 5 \times 2 = 0, \quad 5 \times 3 = 5.$

The pattern continues, so every product $5 \times n$ is $0$ or $5$ and no member multiplies $5$ up to $1$. The same is true of $2$, whose multiples are all even here and so never $1$.

The failure is not universal, and that is worth noticing: $3 \times 7 = 21 \to 1$, so $3$ does have a reciprocal in this system. The proof's hypothesis is that EVERY nonzero member has one, and a single member without one is enough to sink it.

The half that survives. That $3$ is a solution is shown by substituting it: the first factor becomes $3 - 3 = 0$, and $0$ times anything is $0$. Nothing in that uses reciprocals, or division, or any privilege of the real numbers, and the ten-symbol system was built to keep the ordinary rules of expanding and rearranging. So this half survives untouched, and part A confirms it, since $3$ and $6$ are still among the solutions there.

The half that is destroyed. That nothing ELSE is a solution cannot be shown by substituting, because there are always numbers nobody has tried. Over the reals it is the Zero Product Property that closes the gap: if the product of $x - 3$ and $x - 6$ is zero, then one of those two factors is zero, so $x$ is $3$ or $6$ and there is no third possibility. Part B exhibits nonzero members of the ten-symbol system whose product is zero, so that inference is simply unavailable here, and two further members walk through the hole it leaves.

What follows. Reading the zeros off a factored form is really two claims, and only the cheap one is about the factors. The expensive one, that the list is complete, is a claim about the number system the coefficients and the unknown live in. Over the real numbers it holds, so with $a \neq 0$

$a(x - r_1)(x - r_2) = 0 \iff x = r_1 \ \text{ or } \ x = r_2,$

and factoring is a method for SOLVING an equation. In a system with nonzero members multiplying to zero there is no such guarantee: the factorization above still spots the zeros it displays, and part A shows that it misses two others. (A guarantee is exactly what is lost, not every instance of it: $x \cdot x = 0$ has only $x = 0$ in the ten-symbol system too. What you can no longer do is rely on the list being complete without checking.) That is why the lesson proved the property instead of assuming it: the property is the licence, and the licence has a hypothesis.

Over the reals the difference-of-squares template applies as soon as both terms are squares, and it does not insist that the second one be a square of a whole number. Here $4x^2 = (2x)^2$, and $11 = \left(\sqrt{11}\right)^2$ because $\sqrt{11}$ is the nonnegative real number whose square is $11$:

$4x^2 - 11 = (2x)^2 - \left(\sqrt{11}\right)^2.$

$4x^2 - 11 = (2x - \sqrt{11})(2x + \sqrt{11}).$

Expanding confirms it: the cross terms are $2x\sqrt{11}$ and $-2x\sqrt{11}$, which cancel, leaving $4x^2 - 11$.

Now the rational question. This lesson proved that a quadratic with rational coefficients factors into linear factors with rational coefficients exactly when it has a rational zero. So suppose $4x^2 - 11$ had such a factorization. Then it would have a rational zero $r$, and

$4r^2 = 11 \quad \Longrightarrow \quad (2r)^2 = 11.$

So $2r$ would be a rational number whose square is $11$. But $11$ is not a perfect square, so its square root is irrational, while twice a rational number is rational. That is a contradiction, and no rational factorization exists.

Setting the two real factors equal to zero says which numbers the zeros actually are, $\pm\tfrac{\sqrt{11}}{2}$, and they are irrational for the same reason.

Expand the right-hand side, using the perfect-square template with $3x$ in place of the first term and $2$ in place of the second:

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

The identity holds. Now read the sign off it. Whatever real number $x$ is, $3x - 2$ is a real number, and a square is never negative, so

$9x^2 - 12x + 8 = (3x - 2)^2 + 4 \geq 4.$

A quantity that is always at least $4$ is never $0$. So $9x^2 - 12x + 8$ has no real zero.

What that settles is fixed by this lesson's exchange between zeros and factors, and the direction matters. A factorization into two linear factors with real coefficients would hand over a real zero: set the first factor equal to zero and solve it. There is no real zero, so there is no such factorization. That also disposes of the rational case, since a rational zero would be a real one.

What it does not settle is anything outside the system just named. The verdict is about the real numbers, which is the supply of coefficients this question fixed; a later chapter enlarges the number system again, and the same expression has to be asked about afresh there. A verdict is only ever as wide as the system it names.

The direction that holds. Suppose $ax^2 + bx + c$ has integer coefficients and an integer zero $r$. The identity from this lesson gives

$ax^2 + bx + c = (x - r)(ax + b + ar),$

and $a$, $b$ and $r$ are integers, so $b + ar$ is an integer too. Both factors carry integer coefficients. An integer zero really does buy an integer factorization.

The direction that fails. The factorization printed in the prompt is a product of two linear factors with integer coefficients, and expanding checks it: $(3x - 1)(2x + 3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3$. Its zeros come from the Zero Product Property:

$3x - 1 = 0 \ \text{ or } \ 2x + 3 = 0, \qquad x = \tfrac{1}{3} \ \text{ or } \ x = -\tfrac{3}{2}.$

Neither is an integer. So here is a quadratic that factors over the integers and has no integer zero at all, and the claim's "exactly when" is false. What goes wrong is the leading coefficient: the identity puts the whole of $a$ inside the second factor, so a zero of that factor need only be rational. Replace "integer" by "rational" throughout, or by "real" throughout, and both directions do hold, which is what the lesson proved.

The three verdicts, each naming its system.

$4x^2 - 11$ does not factor into linear factors over the rationals, and does factor over the reals.

$9x^2 - 12x + 8$ does not factor into linear factors over the rationals, and does not over the reals either.

$6x^2 + 7x - 3$ factors into linear factors over the integers, and therefore over the rationals and over the reals as well.

The word. What empties "one of these does not factor" is "factor" left bare. The verdict on $4x^2 - 11$ reverses between the rationals and the reals, and over the rationals two of the three fail while over the reals only one does, so without a system the remark does not even pick out which expression is meant, let alone say anything true about it.