A cubic has zeros at $x = -2$, $x = 1$ and $x = 3$, and its $y$-intercept is $12$. Find its equation in factored form.

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

A degree 4 polynomial touches the $x$-axis at $x = -2$ without crossing, crosses it at $x = 1$ and $x = 3$, and passes through $(2, -32)$. Find its equation.

$f(x) = 2(x + 2)^2(x - 1)(x - 3)$

A quartic has a double zero at $x = 0$, single zeros at $x = -\frac{1}{2}$ and $x = \frac{3}{2}$, and passes through $(2, 20)$. Find its equation with integer coefficients.

$f(x) = x^2(2x + 1)(2x - 3) = 4x^4 - 4x^3 - 3x^2$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Finding the Equation of a Polynomial

Every polynomial with real zeros can be written as $f(x) = a(x - r_1)(x - r_2)\cdots$ , and a graph hands you almost all of it. The zeros come from the $x$ -intercepts, the multiplicities from how the graph behaves at each one, and one more point fixes $a$ .

How It Works

Start with the $x$ -intercepts. A zero at $x = r$ is a factor $(x - r)$ , so a zero at $x = -2$ gives $(x + 2)$ , not $(x - 2)$ . Fractions are fine too: a zero at $x = \frac{1}{2}$ gives $(x - \frac{1}{2})$ , or $(2x - 1)$ once you absorb the $2$ into $a$ .

Then read the multiplicity from the shape. The graph cuts straight through the axis at a zero of multiplicity 1, bounces off it at multiplicity 2 (an even power), and flattens out as it passes through at multiplicity 3 (an odd power above 1). The degree is the sum of the multiplicities, so count them and check the total against the end behaviour.

Now you have $f(x) = a(x - r_1)^{m_1}(x - r_2)^{m_2}\cdots$ with one unknown. Substitute any other point on the graph, usually the $y$ -intercept because it is easiest to read, and solve for $a$ . Without that point you only know the family: $x(x - 2)$ and $5x(x - 2)$ have the same zeros.

If the question wants expanded form, expand last, after $a$ is known. Multiply two factors at a time and collect like terms. If it wants integer coefficients and your zeros are fractions, clear them by pairing each fraction with part of the $a$ out front.

Check: substitute the given point back in, and make sure the sign of the leading coefficient matches the end behaviour you can see.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
A cubic has zeros at $x = -2$ , $x = 1$ and $x = 3$ , and its $y$ -intercept is $12$ . Find its equation in factored form.
Show solution
1. Each zero is a factor: $f(x) = a(x + 2)(x - 1)(x - 3)$ .
2. The $y$ -intercept is the point $(0, 12)$ : $f(0) = a(2)(-1)(-3) = 6a$ , so $6a = 12$ and $a = 2$ .
Answer: $f(x) = 2(x + 2)(x - 1)(x - 3)$
Easy
A quartic crosses the $x$ -axis at $x = -3$ and $x = 0$ , bounces off it at $x = 2$ , and passes through $(1, -8)$ . Find its equation.
Show solution
1. Crossing means multiplicity 1 and bouncing means multiplicity 2: $f(x) = ax(x + 3)(x - 2)^2$ . The degree is $1 + 1 + 2 = 4$ , a quartic, as required.
2. Substitute $(1, -8)$ : $f(1) = a(1)(4)(1)^2 = 4a = -8$ , so $a = -2$ .
Answer: $f(x) = -2x(x + 3)(x - 2)^2$
Medium
Find, in expanded form, the cubic with zeros $-1$ , $1$ and $4$ whose graph passes through $(0, 8)$ .
Show solution
1. Factored form: $f(x) = a(x + 1)(x - 1)(x - 4)$ .
2. At $(0, 8)$ : $f(0) = a(1)(-1)(-4) = 4a = 8$ , so $a = 2$ .
3. Expand: $(x + 1)(x - 1) = x^2 - 1$ , then $2(x^2 - 1)(x - 4) = 2(x^3 - 4x^2 - x + 4)$ .
Answer: $f(x) = 2x^3 - 8x^2 - 2x + 8$
Medium
A degree 4 polynomial touches the $x$ -axis at $x = -2$ without crossing, crosses it at $x = 1$ and $x = 3$ , and passes through $(2, -32)$ . Find its equation.
Show solution
1. Touching without crossing is a double zero: $f(x) = a(x + 2)^2(x - 1)(x - 3)$ . Degree $2 + 1 + 1 = 4$ checks.
2. Substitute $(2, -32)$ : $f(2) = a(4)^2(1)(-1) = -16a = -32$ , so $a = 2$ .
Answer: $f(x) = 2(x + 2)^2(x - 1)(x - 3)$
Test level
A quartic has a double zero at $x = 0$ , single zeros at $x = -\frac{1}{2}$ and $x = \frac{3}{2}$ , and passes through $(2, 20)$ . Find its equation with integer coefficients.
Show solution
1. Factored form: $f(x) = ax^2(x + \frac{1}{2})(x - \frac{3}{2})$ .
2. Substitute $(2, 20)$ : $f(2) = a(4)(\frac{5}{2})(\frac{1}{2}) = 5a = 20$ , so $a = 4$ .
3. Clear the fractions by splitting $a = 4 = 2 \times 2$ across the two fraction factors: $4(x + \frac{1}{2})(x - \frac{3}{2}) = (2x + 1)(2x - 3)$ , so $f(x) = x^2(2x + 1)(2x - 3)$ .
4. Expand: $(2x + 1)(2x - 3) = 4x^2 - 4x - 3$ , then multiply by $x^2$ .
Answer: $f(x) = x^2(2x + 1)(2x - 3) = 4x^4 - 4x^3 - 3x^2$

Learn It on Martian Lab

Martian Lab teaches this in the Sketching Polynomials lesson of its MHF4U Polynomial Functions unit. Watch it, then take the Polynomial Functions practice tests to see if it stuck.

Watch the Sketching Polynomials lesson free

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Free Polynomial Functions practice test, with solutions

More Polynomial Functions topics: End Behaviour and Degree · Even and Odd Functions

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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