Graphing Polynomials

Graph polynomials ; a step by step tutorial with examples and detailed solutions. Factoring, zeros and their multiplicities, intercepts and other properties are used to graph polynomials.

Examples with Detailed Solutions

Example 1

Solution to Example 1

• a) Factor $P$ as follows: $$P(x)=-x^3-x^2+2x$$ $$=-x(x^2+x-2)$$ $$=-x(x+2)(x-1)$$
• b) $P$ has three zeros which are $-2$, $0$, and $1$, all with multiplicity one.
• c) The three zeros of $P$ split the number line into four intervals: $$(-\mathrm{\infty },-2),(-2,0),(0,1),(1,+\mathrm{\infty })$$ Select one value of $x$ within each interval and evaluate $P$ at that value to determine the sign of $P$.
$$P(-3)=12>0,P(-1)=-2<0,P\left(\frac{1}{2}\right)=\frac{5}{8}>0,P(2)=-8<0$$ Using the signs of $P$ in each interval, the sign chart is as follows:

Example 2

Solution to Example 2

• a) Factor $P$ as follows $$P(x)=x^4-2x^2+1$$ $$=(x^2-1{)}^2$$ $$=((x-1)(x+1){)}^2$$ $$=(x-1{)}^2(x+1{)}^2$$
• b) Polynomial $P$ has zeros at $x=1$ and $x=-1$, both with multiplicity 2.
• c) Polynomial $P(x)$ is a perfect square and therefore nonnegative for all real values of $x$. $P(x)$ is equal to zero at the two zeros $-1$ and $1$, and positive everywhere else. The sign chart is as follows:

  

• d) The $x$-intercepts are at $(-1,0)$ and $(1,0)$, and the $y$-intercept is at $(0,1)$. The graph of $P$ touches the $x$-axis at $x=-1$ and $x=1$, and opens upward since $P(x)$ is positive, cutting the $y$-axis at $(0,1)$. See graph below:

  

• e) Using the graph of $P$ above, the range of $P$ is given by the interval $$[0,+\mathrm{\infty })$$

Example 3

c) Factor P and determine the multiplicity of each zero of $P$.

d) Determine the sign chart of $P$.

e) Graph polynomial P and label the x and y intercepts on the graph obtained.

Solution to Example 3

• a) Calculate $P(-3)$ $$P(-3)=(-3{)}^4+5(-3{)}^3+5(-3{)}^2-5(-3)-6=0$$ Hence $-3$ is a zero of $P$, and $x+3$ is a factor of $P(x)$.
• b) Since $x+3$ is a factor of $P(x)$, the division of $P(x)$ by $x+3$ must give a remainder equal to zero. Hence $$\frac{P(x)}{x+3}=x^3+2x^2-x-2$$ We now divide $x^3+2x^2-x-2$ by $x-1$: $$\frac{x^3+2x^2-x-2}{x-1}=x^2+3x+2$$ The remainder is 0, which proves that $x-1$ is a factor of $x^3+2x^2-x-2$, and therefore also of $P(x)$.

• c) Using the above, $P(x)$ may be written as: $$P(x)=(x+3)(x-1)(x^2+3x+2)$$ We now factor the quadratic term $x^2+3x+2$ included in $P(x)$. Hence: $$P(x)=(x+3)(x-1)(x+1)(x+2)$$ $P$ has zeros at $x=-3$, $-2$, $-1$, and $1$, and all are of multiplicity one.
• d) The sign chart is shown below:

  

• e) Using the information on the zeros and the sign chart, the graph of $P$ is as shown below with $x$- and $y$-intercepts labeled.

  

Example 4

Solution to Example 4

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