Algebra 1 · A.SSE.B.3A

Factored form

A quadratic can be written so that the places where its graph meets the x-axis are printed in the equation. What else does that form tell you, and what does it hide?

You need

Vertex form: reading, converting and graphingAlgebra 1 Roots, zeros and x-interceptsAlgebra 1

Unlocks

Building a polynomial from its zerosAlgebra 2 Reading a quadratic in its three formsAlgebra 1

Where you are 12 in Quadratics

Algebra9

Quadratics12

▴ 7 above The quadratic formula4 Vertex form: reading, converting and graphing

Factored form

Maximum, minimum and projectile problems Fitting a quadratic to data

Why this exists

Here is one quadratic function written twice:

\begin{aligned} f(x) &= x^2 - 4x - 5 \\ f(x) &= (x + 1)(x - 5) \end{aligned}

Multiply out the second and you get the first, so they are the same function with the same graph. But ask where the graph meets the x-axis and only one of them answers at once. From the first you have to factor, or reach for the quadratic formula. From the second you read it off: the output is 0 when x = -1 and when x = 5, because a product is zero exactly when one of its factors is.

A quadratic can be written in three standard ways, and each one puts a different feature of the parabola in plain sight. Vertex form shows the turning point. Standard form shows where the graph crosses the y-axis. Factored form shows the zeros, and, as it turns out, a great deal more besides.

Definition

A quadratic is in factored form when it is written

f(x) = a(x - r)(x - s) \qquad (a \neq 0)

where a, r and s are numbers. It is a number times two linear factors.

r and s are the zeros of f: the inputs that give the output 0. The graph meets the x-axis at (r, 0) and (s, 0).
a is the same number as the leading coefficient in standard form, the coefficient of x^2. Its sign says which way the parabola opens: upward if a > 0, downward if a < 0.

As with vertex form, the minus signs inside the brackets are there on purpose, so that the numbers you read off are the zeros with their signs intact. A factor with a plus has to be read through that minus:

(x + 1) = \big(x - (-1)\big)

so (x + 1)(x - 5) has zeros -1 and 5. When r = s the two factors are the same and the form becomes a(x - r)^2, a parabola with a single zero that touches the axis at (r, 0).

A product such as (2x + 1)(x - 3) is factored too, just not yet in this exact shape. Taking the 2 out of the first factor gives \displaystyle 2\left(x + \frac{1}{2}\right)(x - 3), and now the zeros \displaystyle -\frac{1}{2} and 3 and the leading coefficient 2 can all be read directly.

Intuition

Look at the graph of y = (x + 1)(x - 5).

Figure 1Everything on this graph can be read from y = (x + 1)(x - 5) with almost no arithmetic.

The two zeros are the obvious part. The rest follows from one more fact: a parabola is symmetric about a vertical line through its vertex, the axis of symmetry. The two x-intercepts are mirror images across that line, so the line sits exactly halfway between them, at the average of the zeros:

x = \frac{-1 + 5}{2} = 2

The vertex is on the axis, so its x-coordinate is 2, and its height is the output there: (2 + 1)(2 - 5) = 3 \cdot (-3) = -9. The y-intercept is the output at 0: (0 + 1)(0 - 5) = -5. Zeros, axis, vertex and y-intercept, all from one line of algebra.

Now keep the zeros and change a. Every function y = a(x - 2)(x - 6) is zero at 2 and at 6, whatever a is, because a times 0 is still 0.

Figure 2y = a(x - 2)(x - 6) for three values of a. The zeros stay put; a decides the rest.

So the zeros pin down two points on the x-axis, and a decides everything else: how far the parabola dips below or rises above the axis, and which way it opens. The axis of symmetry is x = 4 for all three curves; only the height of the vertex changes.

The form also shows where the graph lies below the axis. Take y = (x + 2)(x - 3). Between -2 and 3 the factor x + 2 is positive and x - 3 is negative, so their product is negative. Outside that stretch the two factors have the same sign, and the product is positive.

Figure 3The graph of y = (x + 2)(x - 3) dips below the axis exactly between its zeros.

The zeros are the only places where the sign can change.

Results

What factored form shows. For f(x) = a(x - r)(x - s) with a \neq 0:

the zeros are r and s, and there are no others;
the parabola opens upward if a > 0 and downward if a < 0;
the axis of symmetry is \displaystyle x = \frac{r + s}{2};
the y-intercept is (0, \, ars).

The vertex sits on the axis, at

\left(\frac{r + s}{2},\; -\frac{a(s - r)^2}{4}\right)

From factored form to standard form. Multiplying out,

\begin{aligned} &a(x - r)(x - s) \\ &\quad= ax^2 - a(r + s)x + ars \end{aligned}

so in ax^2 + bx + c, the coefficient b is -a(r + s) and the constant c is ars.

The sign of the output. Take r < s. If a > 0, then f(x) < 0 exactly when r < x < s, and f(x) > 0 when x < r or x > s. If a < 0 the signs are the other way round.

When factored form exists. A quadratic can be written as a(x - r)(x - s) with real numbers r and s exactly when it has real zeros. Those zeros may be whole numbers, fractions or irrational numbers.

Why it is true

The zeros. f(r) = a \cdot 0 \cdot (r - s) = 0 and f(s) = a(s - r) \cdot 0 = 0. For any other x, all three of a, x - r and x - s are non-zero, and a product of non-zero numbers is non-zero. This is the zero product property.

Standard form. Expand the two brackets first, then multiply by a:

\begin{aligned} (x - r)(x - s) &= x^2 - sx - rx + rs \\ &= x^2 - (r + s)x + rs \end{aligned}

The leading coefficient of a(x - r)(x - s) is therefore a, and so the direction of opening is decided by the sign of a exactly as it is in standard form.

The vertex. Put r + s - x in place of x:

\begin{aligned} f(r + s - x) &= a(s - x)(r - x) \\ &= a(x - r)(x - s) = f(x) \end{aligned}

The inputs x and r + s - x always give the same output, and they are mirror images across their average, \displaystyle \frac{r + s}{2}. So the graph is symmetric about \displaystyle x = \frac{r + s}{2}, and the vertex lies on that line. Call it m. Then \displaystyle m - r = \frac{s - r}{2} and \displaystyle m - s = -\frac{s - r}{2}, so

f(m) = a \cdot \frac{s - r}{2} \cdot \left(-\frac{s - r}{2}\right) = -\frac{a(s - r)^2}{4}

The sign. With r < s, the factor x - r is negative left of r and positive right of it, and x - s changes sign at s. Left of both, both factors are negative and their product is positive. Between them, x - r > 0 and x - s < 0, so the product is negative. Right of both, both are positive. Multiplying by a positive a keeps these signs; a negative a flips all three.

When it exists. If f has a real zero r, then (x - r) is a factor of f(x), as shown in roots, zeros and x-intercepts. The other factor is linear, so it has a zero too, and that gives s. If f has no real zeros, then no real r and s can work, because a(x - r)(x - s) would be zero at r.

Worked examples

Example 1

Reading everything off the form

For \displaystyle f(x) = -\frac{1}{2}(x - 2)(x + 4), find the zeros, the direction the parabola opens, the axis of symmetry, the vertex and the y-intercept.

Solution

The zeros are the numbers that make a factor zero. x - 2 = 0 at x = 2, and x + 4 = 0 at x = -4, so the zeros are 2 and -4. The graph meets the x-axis at (2, 0) and (-4, 0).

\displaystyle a = -\frac{1}{2} is negative, so the parabola opens downward and its vertex is the highest point.

The axis is halfway between the zeros:

x = \frac{2 + (-4)}{2} = -1

The vertex lies on the axis, so its height is the output at -1:

f(-1) = -\frac{1}{2}(-3)(3) = \frac{9}{2}

The vertex is \displaystyle \left(-1, \frac{9}{2}\right). The y-intercept is the output at 0: \displaystyle f(0) = -\frac{1}{2}(-2)(4) = 4, the point (0, 4).

Check. In standard form \displaystyle f(x) = -\frac{1}{2}x^2 - x + 4. Its constant term is 4, the same y-intercept, and the vertex formula gives \displaystyle x = -\frac{-1}{2\left(-\frac{1}{2}\right)} = -1, the same axis.

Example 2

From factored form to standard form

Write g(x) = -3(x + 2)(x - 4) in standard form.

Solution

Multiply the two brackets first, since that is a binomial times a binomial, and leave the -3 until the end:

\begin{aligned} (x + 2)(x - 4) &= x^2 - 4x + 2x - 8 \\ &= x^2 - 2x - 8 \end{aligned}

Now distribute the -3 over all three terms:

\begin{aligned} g(x) &= -3(x^2 - 2x - 8) \\ &= -3x^2 + 6x + 24 \end{aligned}

Check. The zeros are r = -2 and s = 4 with a = -3. The result above says b = -a(r + s) = 3 \cdot 2 = 6 and c = ars = -3 \cdot (-2) \cdot 4 = 24. Both agree.

Example 3

From standard form to factored form

Write f(x) = 2x^2 + x - 15 in factored form, and find its vertex.

Solution

The leading coefficient is 2, so factor as for a trinomial whose leading coefficient is not one. We need two numbers whose product is 2 \cdot (-15) = -30 and whose sum is 1. They are 6 and -5, so split the middle term and group:

\begin{aligned} 2x^2 + x - 15 &= 2x^2 + 6x - 5x - 15 \\ &= 2x(x + 3) - 5(x + 3) \\ &= (2x - 5)(x + 3) \end{aligned}

To reach the exact form a(x - r)(x - s), take the 2 out of the first factor, because \displaystyle 2x - 5 = 2\left(x - \frac{5}{2}\right):

f(x) = 2\left(x - \frac{5}{2}\right)(x + 3)

Now a = 2 and the zeros are \displaystyle \frac{5}{2} and -3. The axis is their average,

x = \frac{\frac{5}{2} + (-3)}{2} = -\frac{1}{4}

and the height there is

\begin{aligned} f\left(-\tfrac{1}{4}\right) &= 2 \cdot \left(-\frac{11}{4}\right) \cdot \frac{11}{4} \\ &= -\frac{121}{8} \end{aligned}

The vertex is \displaystyle \left(-\frac{1}{4}, -\frac{121}{8}\right).

Example 4

Three forms of one parabola

Write y = x^2 + 2x - 24 in factored form and in vertex form, and say what each of the three forms shows at a glance.

Solution

For factored form, find two numbers with product -24 and sum 2. They are 6 and -4:

y = (x + 6)(x - 4)

For vertex form, complete the square. Half of 2 is 1, and 1^2 = 1, so add and subtract 1:

\begin{aligned} y &= (x^2 + 2x + 1) - 1 - 24 \\ &= (x + 1)^2 - 25 \end{aligned}

The same parabola, three ways:

standard form x^2 + 2x - 24 shows the y-intercept, (0, -24);
factored form (x + 6)(x - 4) shows the zeros, -6 and 4;
vertex form (x + 1)^2 - 25 shows the vertex, (-1, -25).

All three have leading coefficient 1, since a is the same number in every form. And they agree with each other: the axis x = -1 is halfway between -6 and 4, and (-1 + 6)(-1 - 4) = -25 is the height of the vertex.

Example 5

Zeros that are not whole numbers

Write f(x) = x^2 - 6x + 7 in factored form.

Solution

No two integers multiply to 7 and add to -6: the only pairs, 1 and 7 or -1 and -7, add to 8 and -8. That does not mean there is no factored form. It means the zeros are not integers. Find them by completing the square:

\begin{aligned} x^2 - 6x + 7 &= 0 \\ (x - 3)^2 &= 2 \\ x &= 3 \pm \sqrt{2} \end{aligned}

Each zero gives a factor, and a = 1:

\begin{aligned} f(x) &= \big(x - (3 + \sqrt{2})\big) \\ &\quad \cdot \big(x - (3 - \sqrt{2})\big) \end{aligned}

Check. Group each factor as \displaystyle (x - 3) \mp \sqrt{2}. The product is a difference of squares: \displaystyle (x - 3)^2 - (\sqrt{2})^2 = x^2 - 6x + 9 - 2 = x^2 - 6x + 7.

The axis of symmetry is still the average of the zeros, and the \displaystyle \sqrt{2} cancels: \displaystyle \frac{(3 + \sqrt{2}) + (3 - \sqrt{2})}{2} = 3.

Example 6

A kick, from where it lands

A ball is kicked from the ground and lands 40 m away, and at its highest it is 12 m up. Model its path as a parabola, and find how far from the kick the ball is 9 m up.

Solution

Put the ground on the x-axis with the kick at x = 0 and the landing at x = 40, measuring horizontal distance in meters. The height is 0 at both ends, so 0 and 40 are the zeros, and factored form gives the model straight away:

h(x) = a \cdot x(x - 40)

The highest point is on the axis, halfway between the zeros, at x = 20. Its height is 12:

h(20) = a \cdot 20 \cdot (-20) = -400a = 12

so \displaystyle a = -\frac{3}{100}. It is negative, as it must be for a path that rises and comes back down.

For a height of 9, set h(x) = 9 and multiply both sides by \displaystyle -\frac{100}{3}:

\begin{aligned} x(x - 40) &= -300 \\ x^2 - 40x + 300 &= 0 \\ (x - 10)(x - 30) &= 0 \end{aligned}

So x = 10 or x = 30: the ball is 9 m up at 10 m from the kick on the way up, and again at 30 m on the way down. The two answers are mirror images across x = 20, as the symmetry says they must be.

Check. \displaystyle h(10) = -\frac{3}{100} \cdot 10 \cdot (-30) = 9.

Boundaries

Not every quadratic has a real factored form. x^2 + 2x + 6 = (x + 1)^2 + 5 is never less than 5, so it has no real zeros and cannot be written as a(x - r)(x - s) with real r and s. Whether a quadratic has real zeros can be told in advance from the discriminant. Over the complex numbers it does factor, since it has the two zeros \displaystyle -1 \pm \sqrt{5}\,i, but those zeros are not points on the real graph.

Integers are not required. Factored form exists whenever real zeros exist, including fractional and irrational ones, as in Examples 3 and 5. Failing to factor with whole numbers is a fact about the numbers, not about the form.

The zeros read off only in the exact shape. In (2x - 6)(x + 1) the first factor is zero at x = 3, not at 6. Writing it as 2(x - 3)(x + 1) puts the zero and the leading coefficient where they can be read.

The form hides the vertex height and the y-intercept. Both take a line of arithmetic to get, where vertex form and standard form print them. Each form trades one feature for another; none shows everything.

Where this goes next

Choosing which of the three forms to use for a question, and moving between them, is reading a quadratic in its three forms. Solving an equation that already arrives factored is solving an already-factored quadratic, and the sign result above is the engine of quadratic inequalities.

Running the idea backwards, from chosen zeros and one extra point to a function, is building a polynomial from its zeros. For polynomials of any degree, the link between a zero and a factor is the factor theorem, and a repeated factor like (x - r)^2 is where multiplicity begins: it is why the graph touches the axis instead of crossing it.

Exercises

Reading the form

For each function, find the zeros, the axis of symmetry, the vertex and the y-intercept, and say which way the parabola opens.

f(x) = (x - 4)(x - 10)
g(x) = -(x + 1)(x - 7)
h(x) = 2x(x - 6)
\displaystyle k(x) = \frac{1}{2}(x - 3)(x + 5)
p(x) = -3(x + 1)^2

Factored form to standard form

(x - 3)(x + 8)
-2(x + 1)(x + 4)
\displaystyle 4\left(x - \frac{1}{2}\right)(x - 3)

Standard form to factored form

Write each quadratic in the form a(x - r)(x - s), or explain why it cannot be done with real numbers.

x^2 - 9x + 20
3x^2 - 27
-x^2 + 2x + 24
2x^2 + 7x + 3
x^2 - 4x + 1
x^2 + 6x + 11

Building and using the form

Find the quadratic function whose zeros are -2 and 6 and whose graph passes through (2, -8). Give it in factored form.
A quadratic has zeros 1 and 9, and its largest value is 32. Find it in factored form.

Figure 4The graph for Exercise 17.

Figure 4 shows the graph of a quadratic function f. Write f(x) in factored form, then find its vertex.
The quadratic f(x) = a(x - 3)(x - 9) has y-intercept (0, -27). Find a and the vertex.
A quadratic has leading coefficient 1 and zeros 3 and -7. Use b = -a(r + s) and c = ars to write it in standard form without multiplying out any brackets.

Reasoning

Two quadratic functions both have zeros 2 and 6. Must they be the same function? Explain what can be different about them, and what cannot.
Explain why changing a in a(x - r)(x - s) can move the vertex up or down but never sideways.
Explain why x^2 + 1 cannot be written as a(x - r)(x - s) with real numbers r and s.
Show that the graph of a(x - r)(x - s) meets the y-axis at (0, ars). Use this to explain why a parabola that opens upward and has one negative zero and one positive zero must cross the y-axis below the origin.
Explain why (2x - 6)(x + 1) and 2(x - 3)(x + 1) have the same zeros, and say what the 2 in the second one tells you that the first does not show directly.

Answers

Zeros 4 and 10; axis x = 7; vertex (7, -9); y-intercept (0, 40); opens upward. \displaystyle \frac{4 + 10}{2} = 7 and f(7) = 3 \cdot (-3) = -9; f(0) = (-4)(-10) = 40.
Zeros -1 and 7; axis x = 3; vertex (3, 16); y-intercept (0, 7); opens downward. g(3) = -(4)(-4) = 16 and g(0) = -(1)(-7) = 7; a = -1 is negative.
Zeros 0 and 6; axis x = 3; vertex (3, -18); y-intercept (0, 0); opens upward. The factor x is x - 0, and h(3) = 2 \cdot 3 \cdot (-3) = -18.
Zeros 3 and -5; axis x = -1; vertex (-1, -8); y-intercept \displaystyle \left(0, -\frac{15}{2}\right); opens upward. \displaystyle k(-1) = \frac{1}{2}(-4)(4) = -8 and \displaystyle k(0) = \frac{1}{2}(-3)(5) = -\frac{15}{2}.
One zero, -1; axis x = -1; vertex (-1, 0); y-intercept (0, -3); opens downward. Here r = s = -1, so the vertex is the zero itself, on the x-axis, and p(0) = -3 \cdot 1 = -3.
x^2 + 5x - 24. x^2 + 8x - 3x - 24.
-2x^2 - 10x - 8. (x + 1)(x + 4) = x^2 + 5x + 4, then multiply by -2.
4x^2 - 14x + 6. \displaystyle \left(x - \frac{1}{2}\right)(x - 3) = x^2 - \frac{7}{2}x + \frac{3}{2}, then multiply by 4.
(x - 4)(x - 5). -4 and -5 multiply to 20 and add to -9.
3(x - 3)(x + 3). Take out the 3: 3(x^2 - 9), a difference of squares.
-(x - 6)(x + 4). Take out -1: -(x^2 - 2x - 24), and -6 and 4 multiply to -24 and add to -2.
\displaystyle 2\left(x + \frac{1}{2}\right)(x + 3). 2x^2 + 7x + 3 = (2x + 1)(x + 3), and \displaystyle 2x + 1 = 2\left(x + \frac{1}{2}\right).
\displaystyle \big(x - (2 + \sqrt{3})\big)\big(x - (2 - \sqrt{3})\big). Completing the square, (x - 2)^2 = 3, so the zeros are \displaystyle 2 \pm \sqrt{3}.
It cannot be done with real numbers. x^2 + 6x + 11 = (x + 3)^2 + 2 \geq 2, so there are no real zeros.
\displaystyle f(x) = \frac{1}{2}(x + 2)(x - 6). f(2) = a(4)(-4) = -16a = -8, so \displaystyle a = \frac{1}{2}.
f(x) = -2(x - 1)(x - 9). The largest value is at the vertex, x = 5, and f(5) = a(4)(-4) = -16a = 32, so a = -2.
\displaystyle f(x) = \frac{1}{2}(x + 4)(x - 2), vertex \displaystyle \left(-1, -\frac{9}{2}\right). The zeros are -4 and 2, and f(0) = a(4)(-2) = -8a = -4 gives \displaystyle a = \frac{1}{2}; then \displaystyle f(-1) = \frac{1}{2}(3)(-3) = -\frac{9}{2}.
a = -1, vertex (6, 9). f(0) = a(-3)(-9) = 27a = -27, so a = -1; the axis is \displaystyle x = \frac{3 + 9}{2} = 6, and f(6) = -(3)(-3) = 9.
x^2 + 4x - 21. b = -(3 + (-7)) = 4 and c = 3 \cdot (-7) = -21.
No. Both are a(x - 2)(x - 6), but a can be any non-zero number. So they share their zeros and their axis, x = 4, but can differ in which way they open, how steep they are, the height of the vertex and the y-intercept. Each is a constant multiple of the other.
The axis of symmetry is \displaystyle x = \frac{r + s}{2}, which does not involve a, so the vertex stays on that line. Only its height, \displaystyle -\frac{a(s - r)^2}{4}, depends on a.
x^2 + 1 \geq 1 for every real x, so it is never 0 and has no real zeros. But a(x - r)(x - s) is 0 at x = r, so the two cannot be equal.
At x = 0 the output is a(0 - r)(0 - s) = a(-r)(-s) = ars. Opening upward means a > 0, and zeros of opposite signs make rs < 0, so ars < 0: the y-intercept is below the origin.
2x - 6 = 2(x - 3), so the two expressions are equal for every x and have the same zeros, 3 and -1. The 2 out front is the leading coefficient: the parabola opens upward, and every output is twice the output of (x - 3)(x + 1).