How to Make a Sign Table for Polynomials
Questions with Solutions
Master Grade 12 math with this step-by-step guide on creating sign tables for polynomial functions. This resource includes challenging practice questions, detailed solutions, and clear graphical interpretations to help you fully understand polynomial behavior.
Core Practice Problems
Question 1 - Creating a Sign Table from Factored Form
Polynomial $p$ is given by:
$$ p(x) = (x - 1)^2(x - \sqrt{3}) (x + \sqrt{3}) $$Make a sign table of $p$ and sketch a possible graph for $p(x)$.
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1. Find the Zeros:
We first find the zeros of the polynomial function $p(x)$ by setting it to zero:
For $p(x) = 0$, we need to have:
$$ (x - 1)^2 = 0 \quad \text{or} \quad (x - \sqrt{3}) = 0 \quad \text{or} \quad (x + \sqrt{3}) = 0 $$Solve each equation to obtain the zeros of $p(x)$:
$$ x = 1 \text{ (with multiplicity 2)}, \quad x = \sqrt{3}, \quad \text{and} \quad x = -\sqrt{3} $$2. Analyze the Factors:
With the help of the factored form and its zeros, we study the sign of each factor:
- $(x - 1)^2$ is always positive for all $x$ except at $x = 1$.
- $x - \sqrt{3} > 0$ for $x > \sqrt{3}$.
- $x + \sqrt{3} > 0$ for $x > -\sqrt{3}$.
3. Construct the Sign Table:
We put each factor in the table and use the rules of multiplication of signs to complete the sign for $p(x)$.
4. Sketch the Graph:
We use the zeros of $p(x)$ (which act as $x$-intercepts), the table of signs, and the $y$-intercept $(0, -3)$ to complete the graph.
Question 2 - Building a Polynomial and Sign Table from Properties
$f(x)$ is a polynomial of degree six with a negative leading coefficient $k$. $f$ has a zero of multiplicity 1 at $x = -1$, a zero of multiplicity 3 at $x = 1$, and a zero of multiplicity 2 at $x = 3$. Make a sign table for the polynomial $f(x)$.
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1. Write the Factors:
We first write the factors of polynomial $f$ along with their respective multiplicities:
- Zero of multiplicity 1 at $x = -1 \implies \text{factor: } (x + 1)$
- Zero of multiplicity 3 at $x = 1 \implies \text{factor: } (x - 1)^3$
- Zero of multiplicity 2 at $x = 3 \implies \text{factor: } (x - 3)^2$
2. Formulate the Equation:
Let $k$ (negative) be the leading coefficient of $f$. Using all the factors above, we write $f(x)$ as:
3. Study the Signs of the Factors:
- $x + 1 > 0$ for $x > -1$
- $(x - 1)^3 > 0$ for $x > 1$
- $(x - 3)^2 > 0$ for all $x$ except $x = 3$
- $k < 0$ means it will flip the final signs of the product.
4. Construct the Sign Table:
Below is the table of signs for each factor and the final polynomial $f(x)$ in the bottom row.
Prepare for Standardized Tests
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Free Practice for SAT, ACT and Compass Maths Tests →Challenge Problems
Push your understanding of sign tables with these advanced scenarios. We use comprehensive KaTeX array tables to map out the sign changes.
Challenge 1 - Factoring Before Making a Sign Table
Find the intervals where the polynomial $g(x) = -2x^5 + 8x^3$ is positive by constructing a sign table.
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1. Factor the Polynomial:
First, factor out the greatest common factor, which is $-2x^3$:
Now, factor the difference of squares:
$$ g(x) = -2x^3(x - 2)(x + 2) $$2. Find the Zeros:
Setting $g(x) = 0$ gives zeros at $x = 0$ (multiplicity 3), $x = 2$ (multiplicity 1), and $x = -2$ (multiplicity 1). These divide the number line into four intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, and $(2, \infty)$.
3. Construct the Sign Table:
Use a LaTeX array to determine the sign of the overall polynomial across the critical intervals.
Conclusion:
By observing the bottom row of the sign table, the polynomial $g(x)$ is positive on the intervals $\mathbf{(-\infty, -2) \cup (0, 2)}$.
Challenge 2 - Factoring by Grouping
Let $h(x) = x^3 - 3x^2 - 4x + 12$. Determine the intervals where $h(x) < 0$.
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1. Factor by Grouping:
Group the first two terms and the last two terms together:
Factor out the common binomial $(x - 3)$:
$$ h(x) = (x^2 - 4)(x - 3) $$Factor the difference of squares:
$$ h(x) = (x - 2)(x + 2)(x - 3) $$2. Find Zeros and Intervals:
The roots are $x = -2, 2, \text{ and } 3$. Since they all have an odd multiplicity of 1, the sign will alternate at every root.
3. Construct the Sign Table:
$$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & (-\infty, -2) & -2 & (-2, 2) & 2 & (2, 3) & 3 & (3, \infty) \\ \hline x+2 & - & 0 & + & + & + & + & + \\ \hline x-2 & - & - & - & 0 & + & + & + \\ \hline x-3 & - & - & - & - & - & 0 & + \\ \hline h(x) & - & 0 & + & 0 & - & 0 & + \\ \hline \end{array} $$Conclusion:
Based on the table, $h(x) < 0$ on the intervals $\mathbf{(-\infty, -2) \cup (2, 3)}$.
Challenge 3 - Irreducible Quadratics and Even Powers
Determine the sign of the polynomial $P(x) = x^4(x^2 + 1)(x - 5)^3$ across all real numbers using a sign table.
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1. Analyze Each Factor:
- $x^2 + 1$: This is an irreducible quadratic. It is strictly positive ($>0$) for all real numbers $x$.
- $x^4$: This is an even power. It is positive for all $x$, except at $x = 0$ where it equals 0.
- $(x - 5)^3$: This is an odd power. It is negative when $x < 5$, zero at $x = 5$, and positive when $x > 5$.
2. Construct the Sign Table:
We place the critical points $x=0$ and $x=5$ into our intervals.
Conclusion:
- $P(x) < 0$ for $\mathbf{x \in (-\infty, 0) \cup (0, 5)}$
- $P(x) = 0$ at $\mathbf{x = 0}$ and $\mathbf{x = 5}$
- $P(x) > 0$ for $\mathbf{x \in (5, \infty)}$