Families of Polynomial Functions

In Grade 11 you saw that many parabolas share the same two zeros, and one extra point picks out a single family member. The same idea works for polynomials of any degree. Given the zeros, you can write a whole family of polynomials; given one more point, you can find the exact equation.

Key ideas

A family with the same zeros

The polynomials with zeros $r_1, r_2, \dots, r_n$ (and degree $n$) form the family

$$y = k(x - r_1)(x - r_2)\cdots(x - r_n), \qquad k \in \mathbb{R},\ k \ne 0$$

Every member has the same $x$-intercepts. Changing $k$ stretches the graph vertically, and a negative $k$ flips it.

Every member of $y = k(x + 2)(x - 1)(x - 3)$ crosses the $x$-axis at $-2$, $1$, and $3$.

Finding one member

1. Write the family from the zeros (with the right multiplicities).
2. Substitute the given point $(x, y)$.
3. Solve for $k$.
4. Write the equation (expand only if asked).

Repeated zeros and degree

A zero of order $m$ gives a factor to the power $m$. The degree is the sum of the orders. For example, a quartic with zeros $-2$ (order $2$), $1$, and $3$ is

$$y = k(x + 2)^2(x - 1)(x - 3)$$

If you’re told only the $x$-intercepts and the degree, there’s usually more than one way to choose the orders, so more than one family fits. Clues like “touches the axis” (even order) or “crosses” (odd order) narrow it down.

Zeros that are fractions

For a zero like $\dfrac{1}{2}$, you can use the factor $(2x - 1)$ instead of $\left(x - \tfrac{1}{2}\right)$. Both are zero at $x = \tfrac{1}{2}$, and the whole-number version is easier to work with. The value of $k$ just adjusts.

Using finite differences

If you have a table of values instead of a graph:

1. Use finite differences to find the degree $n$ and the leading coefficient $a$ (from $a \cdot n!$).
2. Read the zeros from the table (where $y = 0$).
3. If you’ve found $n$ zeros, the equation is $y = a(x - r_1)\cdots(x - r_n)$, because $k$ is the leading coefficient.

Worked examples

Example 1: The member through a point

Write the family of cubic functions with zeros $-2$, $1$, and $3$. Then find the member that passes through $(2, -8)$.

Solution. The family is

$$y = k(x + 2)(x - 1)(x - 3), \qquad k \ne 0$$

Substitute $x = 2$, $y = -8$:

$$\begin{aligned} -8 &= k(2 + 2)(2 - 1)(2 - 3) \\ -8 &= k(4)(1)(-1) \\ -8 &= -4k \\ k &= 2 \end{aligned}$$

The member is $y = 2(x + 2)(x - 1)(x - 3)$.

Example 2: A repeated zero and the y-intercept

A quartic function has a zero of order $2$ at $-2$, zeros of order $1$ at $1$ and $3$, and a $y$-intercept of $24$. Find its equation.

Solution.

$$y = k(x + 2)^2(x - 1)(x - 3)$$

The $y$-intercept means the point $(0, 24)$:

$$24 = k(2)^2(-1)(-3) = 12k \quad\Rightarrow\quad k = 2$$

The equation is $y = 2(x + 2)^2(x - 1)(x - 3)$.

Example 3: More than one answer

Find possible equations for a quartic function whose only zeros are $-1$ and $2$ (all its zeros are real), if the graph crosses the $x$-axis at both.

Solution. Crossing means odd order. The two orders must be odd and add to $4$, so they are $1$ and $3$, in either order:

$$y = k(x + 1)(x - 2)^3 \qquad \text{or} \qquad y = k(x + 1)^3(x - 2)$$

where $k$ is any nonzero number. There are infinitely many quartics that fit: two families, and any $k$ in each. (If the graph touched the axis at both, the orders would be $2$ and $2$: $y = k(x + 1)^2(x - 2)^2$.) Without the “all zeros real” condition there are even more answers, such as $y = k(x + 1)(x - 2)(x^2 + 1)$, because $x^2 + 1$ has no real zeros and adds no $x$-intercepts (see Practice 9).

Example 4: From a table of values

Find an equation for the polynomial function in the table.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$-40$	$0$	$12$	$8$	$0$	$0$

Solution. The $x$-values go up by $1$.

• First differences: $40, 12, -4, -8, 0$
• Second differences: $-28, -16, -4, 8$
• Third differences: $12, 12, 12$

The degree is $3$, and $6a = 12$ gives $a = 2$.

The table shows $y = 0$ at $x = -1$, $2$, and $3$. That’s three zeros for a cubic, so

$$y = 2(x + 1)(x - 2)(x - 3)$$

Check with $(0, 12)$: $2(1)(-2)(-3) = 12$. ✓

Common mistakes

Mixing up the sign of the zero and the factor. A zero of $-2$ gives the factor $(x + 2)$.

Forgetting the order of a repeated zero. “A zero of order $2$ at $-2$” means $(x + 2)^2$, not $(x + 2)$. Without the square, the degree and the graph are both wrong.

Leaving out $k$. $y = (x + 2)(x - 1)(x - 3)$ is just one member. The family needs $k$, and you can’t find the member through a point without it.

Arithmetic slips when substituting. Work out each bracket separately, as in Example 1, before multiplying. A single sign error changes $k$.

Assuming only one polynomial fits. Zeros and a degree alone usually allow many polynomials. You need a point to find $k$, and sometimes extra information (touch or cross) to decide the orders.

Practice

1. (Warm-up) Write the family of cubic functions with zeros $2$, $-3$, and $5$.

Solution

$$y = k(x - 2)(x + 3)(x - 5), \qquad k \ne 0$$

2. (Warm-up) Are these in the same family as $y = 3(x - 1)(x + 2)$? (a) $y = -(x + 2)(x - 1)$ (b) $y = (x - 1)(x + 2)^2$

Solution

(a) Yes. It has the same factors with $k = -1$ (the order of the factors doesn’t matter).

(b) No. It has a repeated factor, so it’s a cubic with a different shape at $-2$. It isn’t of the form $k(x - 1)(x + 2)$.

3. (Warm-up) Find the member of the family $y = kx(x - 2)(x + 3)$ that passes through $(1, 8)$.

Solution

$$8 = k(1)(1 - 2)(1 + 3) = -4k \quad\Rightarrow\quad k = -2$$

$y = -2x(x - 2)(x + 3)$.

4. (Core) Find the cubic function with zeros $-2$, $\tfrac{1}{2}$, and $3$ that passes through $(1, -6)$.

Solution

Use the factor $(2x - 1)$ for the zero $\tfrac{1}{2}$:

$$y = k(x + 2)(2x - 1)(x - 3)$$$$-6 = k(3)(1)(-2) = -6k \quad\Rightarrow\quad k = 1$$

$y = (x + 2)(2x - 1)(x - 3)$.

5. (Core) A quartic function has a zero of order $3$ at $1$ and a zero of order $1$ at $-2$, and passes through $(2, 8)$. Find its equation.

Solution

$$y = k(x - 1)^3(x + 2)$$$$8 = k(1)^3(4) = 4k \quad\Rightarrow\quad k = 2$$

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

6. (Core) A quartic function crosses the $x$-axis at $-3$ and $4$, touches it at $0$, has no other $x$-intercepts, and passes through $(1, -24)$. Find its equation.

Solution

Crossing at $-3$ and $4$ (order $1$ each) and touching at $0$ (order $2$) gives degree $1 + 1 + 2 = 4$:

$$y = k(x + 3)x^2(x - 4)$$$$-24 = k(4)(1)(-3) = -12k \quad\Rightarrow\quad k = 2$$

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

7. (Core) Find an equation for the polynomial function in the table.

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$12$	$0$	$-2$	$0$	$0$

Solution

• First differences: $-12, -2, 2, 0$
• Second differences: $10, 4, -2$
• Third differences: $-6, -6$

Degree $3$, and $6a = -6$ gives $a = -1$. The zeros in the table are $1$, $3$, and $4$:

$$y = -(x - 1)(x - 3)(x - 4)$$

Check with $(0, 12)$: $-(-1)(-3)(-4) = 12$. ✓

8. (Challenge) Find the cubic function with zeros $1 + \sqrt{2}$, $1 - \sqrt{2}$, and $3$ whose $y$-intercept is $6$.

Solution

Multiply the two irrational factors with the difference of squares:

$$\big(x - (1 + \sqrt{2})\big)\big(x - (1 - \sqrt{2})\big) = (x - 1)^2 - 2 = x^2 - 2x - 1$$

So $y = k(x^2 - 2x - 1)(x - 3)$. At $(0, 6)$:

$$6 = k(-1)(-3) = 3k \quad\Rightarrow\quad k = 2$$

$y = 2(x^2 - 2x - 1)(x - 3)$.

9. (Challenge) Explain why no polynomial of degree $5$ can cross the $x$-axis at $-3$ and $4$, touch it at $0$, and have no other $x$-intercepts.

Solution

Crossing zeros have odd order and touching zeros have even order. So the orders at $-3$, $0$, and $4$ add up to odd $+$ even $+$ odd, which is even.

Any other factor would have to have no real zeros (like $x^2 + 1$). Such a factor always has even degree, because a polynomial of odd degree always crosses the $x$-axis somewhere.

So the total degree must be even, and it can’t be $5$. (Degree $4$ works: see question 6.)