Polynomials in Factored Form

When a polynomial is written as a product of factors, like $f(x) = 2(x - 3)(x + 1)^2$, its graph is almost drawn for you. Each factor gives an $x$-intercept, the power on the factor tells you how the graph behaves there, and the leading coefficient sets the ends. This page shows how to put those pieces together into a quick, accurate sketch.

Key ideas

Zeros from factors

A product is zero only when one of its factors is zero. So for

$$f(x) = a(x - r_1)(x - r_2)\cdots(x - r_n)$$

the zeros (the $x$-intercepts) are $r_1, r_2, \dots, r_n$. Watch the signs: the factor $x + 4$ gives the zero $x = -4$, and $2x - 1$ gives $x = \dfrac{1}{2}$.

The $y$-intercept is $f(0)$: replace every $x$ with $0$ and multiply.

Degree and leading coefficient

Multiply the leading terms of all the factors (including powers). For $f(x) = 2(x - 3)(x + 1)^2$, the leading term is $2 \cdot x \cdot x^2 = 2x^3$: degree $3$, leading coefficient $2$. Then use end behaviour to see where the ends go.

Multiplicity: what happens at each zero

If a factor $(x - r)$ appears $m$ times, the zero $r$ has multiplicity (or order) $m$. The multiplicity controls the shape of the graph at that $x$-intercept:

Order	Shape at the zero	Sign of $y$
$1$	crosses the axis, like a line	changes
$2$	touches the axis and turns back, like a parabola	stays the same
$3$	crosses but flattens out as it goes through, like $y = x^3$	changes

The same zero, $x = 1$, with order 1, 2, and 3.

In general: an odd order means the graph crosses (the sign changes), and an even order means it touches and turns back (the sign doesn’t change). Higher orders look flatter near the zero.

Sign intervals

The zeros split the $x$-axis into intervals. On each interval, $f(x)$ is either positive (graph above the axis) or negative (graph below). To find out which, either

• pick a test value in each interval and find the sign of $f$ there, or
• start from the end behaviour on the right and move left, changing sign at each odd-order zero and keeping the sign at each even-order zero.

Steps for sketching

1. Find the zeros and their orders.
2. Find the $y$-intercept.
3. Find the degree and leading coefficient, and the end behaviour.
4. Find the sign on each interval.
5. Plot the intercepts and join them with a smooth curve that matches the ends, the signs, and the behaviour at each zero.

Worked examples

Example 1: Reading the features

For $f(x) = 2(x - 3)(x + 1)^2(x + 4)$, state the zeros and their orders, the behaviour at each zero, the degree, the end behaviour, and the $y$-intercept.

Solution.

Zero	Order	Behaviour
$3$	$1$	crosses
$-1$	$2$	touches
$-4$	$1$	crosses

The leading term is $2 \cdot x \cdot x^2 \cdot x = 2x^4$: degree $4$, leading coefficient $2$. The ends both go up (Q2 to Q1).

$$f(0) = 2(-3)(1)^2(4) = -24$$

The $y$-intercept is $-24$.

Example 2: A full sketch

Sketch $f(x) = -(x + 2)(x - 1)^2$, and state where $f(x) \gt 0$ and where $f(x) \lt 0$.

Solution.

1. Zeros: $-2$ (order $1$, crosses) and $1$ (order $2$, touches).
2. $y$-intercept: $f(0) = -(2)(1) = -2$.
3. Leading term: $-x \cdot x^2 = -x^3$. Degree $3$, negative, so Q2 to Q4.
4. Signs, using test values:

Interval	$x \lt -2$	$-2 \lt x \lt 1$	$x \gt 1$
test value	$x = -3$	$x = 0$	$x = 2$
$f(x)$	$-(-1)(16) = 16$	$-2$	$-(4)(1) = -4$
sign	$+$	$-$	$-$

Notice the sign changes at $-2$ (odd order) but not at $1$ (even order).

The graph of $f(x) = -(x + 2)(x - 1)^2$ crosses at $-2$ and touches at $1$.

$f(x) \gt 0$ for $x \lt -2$, and $f(x) \lt 0$ for $-2 \lt x \lt 1$ or $x \gt 1$.

Example 3: Factor first

Find the zeros of $f(x) = x^4 - 5x^2 + 4$, and find where $f(x) \gt 0$.

Solution. This factors like a quadratic in $x^2$:

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

The zeros are $-2$, $-1$, $1$, and $2$, all of order $1$. The leading coefficient is $1$ and the degree is $4$, so the graph is positive at both ends (Q2 to Q1). Moving left from the right end, the sign changes at every zero:

Interval	$x \lt -2$	$-2 \lt x \lt -1$	$-1 \lt x \lt 1$	$1 \lt x \lt 2$	$x \gt 2$
sign	$+$	$-$	$+$	$-$	$+$

Check one: $f(0) = 4 \gt 0$. ✓

$f(x) \gt 0$ for $x \lt -2$, $-1 \lt x \lt 1$, or $x \gt 2$.

Example 4: A factor written backwards

For $h(x) = (2 - x)(x + 3)^3$, find the degree, the leading coefficient, the end behaviour, the $y$-intercept, and the behaviour at each zero.

Solution. The leading term of $2 - x$ is $-x$, not $x$:

$$(-x)(x^3) = -x^4$$

Degree $4$, leading coefficient $-1$: both ends go down (Q3 to Q4).

$y$-intercept: $h(0) = (2)(3)^3 = 54$.

Zeros: $2 - x = 0$ gives $x = 2$ (order $1$, crosses). $x = -3$ has order $3$, so the graph flattens out as it crosses there.

Common mistakes

Getting the sign of a zero wrong. The factor $(x + 4)$ gives the zero $-4$, not $4$. Set each factor equal to $0$ and solve.

Ignoring a negative hidden inside a factor. In $(2 - x)$ or $(1 - 2x)$, the $x$-term is negative, so it flips the leading coefficient. In Example 4, $h$ opens down even though there’s no minus sign in front.

Forgetting the powers when finding the degree or $y$-intercept. In $2(x - 3)(x + 1)^2(x + 4)$, the degree is $1 + 2 + 1 = 4$, and the $y$-intercept uses $(1)^2$.

Crossing at an even-order zero. At a zero of order $2$ (or $4$), the graph touches the axis and turns back. It doesn’t cross, so the sign is the same on both sides.

Drawing an order-3 zero like an order-1 zero. Both cross, but at an order-3 zero the graph flattens out as it passes through, like $y = x^3$ at the origin.

Practice

1. (Warm-up) State the zeros of $f(x) = (x - 5)^2(x + 3)$ and their orders. At which zero does the graph cross the $x$-axis?

Solution

$5$ (order $2$) and $-3$ (order $1$). The graph crosses at $-3$ and touches at $5$.

2. (Warm-up) Find the $y$-intercept of $f(x) = -3(x + 1)(x - 2)(x - 4)$.

Solution

$$f(0) = -3(1)(-2)(-4) = -24$$

3. (Warm-up) Describe the shape of the graph of $y = (x + 4)^3(x - 1)^2$ at each $x$-intercept.

Solution

At $x = -4$ (order $3$), the graph crosses the axis and flattens out as it goes through. At $x = 1$ (order $2$), it touches the axis and turns back.

4. (Core) Find the degree, the leading coefficient, and the end behaviour of $f(x) = (1 - 2x)(x + 3)^2$.

Solution

$$(-2x)(x^2) = -2x^3$$

Degree $3$, leading coefficient $-2$. As $x \to -\infty$, $y \to \infty$, and as $x \to \infty$, $y \to -\infty$: Q2 to Q4.

5. (Core) Find the intervals where $f(x) = (x + 3)(x - 1)(x - 4)$ is positive and where it is negative.

Solution

Zeros $-3$, $1$, $4$, all order $1$. Degree $3$ with a positive leading coefficient, so the right end is positive, and the sign changes at each zero.

Interval	$x \lt -3$	$-3 \lt x \lt 1$	$1 \lt x \lt 4$	$x \gt 4$
sign	$-$	$+$	$-$	$+$

Check: $f(0) = (3)(-1)(-4) = 12 \gt 0$. ✓

Positive for $-3 \lt x \lt 1$ or $x \gt 4$. Negative for $x \lt -3$ or $1 \lt x \lt 4$.

6. (Core) For $f(x) = x(x - 2)^2(x + 1)$, give the zeros and their behaviour, the end behaviour, and the intervals where $f(x) \lt 0$. Then describe the sketch.

Solution

Zeros: $-1$ (order $1$, crosses), $0$ (order $1$, crosses), $2$ (order $2$, touches). Leading term $x \cdot x^2 \cdot x = x^4$, so both ends go up (Q2 to Q1).

Test values:

• $x = -2$: $(-2)(16)(-1) = 32 \gt 0$
• $x = -0.5$: $(-0.5)(6.25)(0.5) \lt 0$
• $x = 1$: $(1)(1)(2) = 2 \gt 0$
• $x = 3$: $(3)(1)(4) = 12 \gt 0$

$f(x) \lt 0$ only for $-1 \lt x \lt 0$.

Sketch: come down from the upper left, cross at $-1$, dip below the axis, come back up through the origin, rise and then fall to touch the axis at $(2, 0)$, and rise again to the upper right.

7. (Core) Factor $f(x) = 2x^3 + 2x^2 - 12x$ fully. Then find where $f(x) \gt 0$.

Solution

$$2x^3 + 2x^2 - 12x = 2x(x^2 + x - 6) = 2x(x + 3)(x - 2)$$

Zeros $-3$, $0$, $2$, all order $1$. Degree $3$, positive leading coefficient: the right end is positive.

Interval	$x \lt -3$	$-3 \lt x \lt 0$	$0 \lt x \lt 2$	$x \gt 2$
sign	$-$	$+$	$-$	$+$

Check: $f(-1) = 2(-1)(2)(-3) = 12 \gt 0$. ✓

$f(x) \gt 0$ for $-3 \lt x \lt 0$ or $x \gt 2$.

8. (Challenge) For $f(x) = -(x + 1)^2(x - 2)^3$, describe the end behaviour and the behaviour at each zero, and find all $x$ where $f(x) \gt 0$.

Solution

Leading term: $-x^2 \cdot x^3 = -x^5$. Degree $5$, negative: Q2 to Q4.

At $x = -1$ (order $2$) the graph touches the axis. At $x = 2$ (order $3$) it crosses and flattens out.

Start on the right, where $f$ is negative. Moving left, the sign changes at $2$ (odd order) and stays the same at $-1$ (even order):

Interval	$x \lt -1$	$-1 \lt x \lt 2$	$x \gt 2$
sign	$+$	$+$	$-$

Check: $f(0) = -(1)(-8) = 8 \gt 0$ ✓ and $f(-2) = -(1)(-64) = 64 \gt 0$ ✓.

$f(x) \gt 0$ for $x \lt -1$ or $-1 \lt x \lt 2$ (that is, every $x \lt 2$ except $x = -1$, where $f = 0$).

9. (Challenge) Explain why $f(x) = (x^2 + 1)(x - 2)$ has degree $3$ but only one $x$-intercept. Where is $f(x)$ negative?

Solution

The factor $x^2 + 1$ is never zero, because $x^2 \ge 0$ means $x^2 + 1 \ge 1$. So the only zero comes from $x - 2 = 0$: $x = 2$. The degree is still $2 + 1 = 3$, but a degree only gives the maximum number of $x$-intercepts.

Since $x^2 + 1$ is always positive, $f(x)$ has the same sign as $x - 2$. So $f(x) \lt 0$ for $x \lt 2$.