Fourth Degree Polynomials

Several fourth degree polynomials are presented along with questions with detailed solutions.

Questions

Question 1
About: Polynomial of the fourth degree: touches the x axis at one point.
Question: Why does the graph touches (but not cut) the x-axis at one point only?

Question 2
About: Polynomial of the Fourth degree: 2 x-intercepts.
Question: If the graph cuts the x axis at x = 1, what are the coordinates of the other x-intercpet?

Question 3
About: Polynomial of the Fourth degree: 3 x-intercepts and parameter $a$ to determine.
Question: The graph below touches (but does not cut) the x-axis at x = 2. What are the coordinates of the other two x-intercpets?

Question 4
About: Polynomial of a fourth degree: no x-intercepts.
Question: Why does the graph of the fourth degree polynomial $y = x^4+x^3+2x^2+x+1$ have no x-intercept knowing that $x^2 + 1$ is a factor of this polynomial?

Answers to the Above Questions

Answer Question 1
Examine the equation of the polynomial given: $y = x^4$. Solve $x^4 = 0$ to obtain a zero of multiplicity 4, hence the the graph touches the x-axis at one point but the graph is flat at $x = 0$ indicating the mutliplicity $4$ .

Answer Question 2
An x intercept at $x = 1$ means that $x - 1$ is a factor of the given polynomial and we can write: $y = x^4+0.5x-x^3-0.5$ = (x-1) Q(x) \).
Using polynomial division to find: $Q(x) = \dfrac{x^4+0.5x-x^3-0.5}{x-1} = x^3+0.5$ and therefore the given polynomial can be written as: $y = (x - 1)(x^3+0.5)$.
Find the other zero(s) by solving $x^3+0.5 = 0$ which has one real solution given by $x = - \sqrt[3]{0.5} \approx - 0.8$ as shown in the graph above.

Answer Question 3
The graph of the polynomial touches the x-axis at $x = 2$ and therefore $y = 0$ for $x = 2$. Hence the equation
$(2)^4-2(2)^3-5(2)^2+ a(2) - 4 = 0$
Simplify and solve for $a$ to obtain $a = 12$
Substitute $a$ by $12$ to write the given polynomial as $y = x^4-2x^3-5x^2+ 12 x-4$
Since the graph touches the x-axis at $x = 2$, $x = 2$ is a zero of an even multiplicity (2, 4, 6,...). The multipliciy cannot be more than $2$ since the maximum number of zeros is no more than the degree of the polynomial which $4$ and the graph has two other x-intercepts.
Therefore the polynomial $y = x^4-2x^3-5x^2+ 12 x-4$ may be written as $y = (x-2)^2 Q(x)$ where $Q(x) = \dfrac{x^4-2x^3-5x^2+ 12 x-4}{(x-2)^2}$ and using polynomial division, we obtain
$Q(x) = x^2+2x-1$
The two remaining zeros are found by solving
$x^2+2x-1 = 0$
which gives the solution $x = -1+\sqrt{2} \approx 0.41$ and $x= - 1 - \sqrt{2} \approx -2.41$ that are shown in the given graph of the polynomial.

Answer Question 4
Since $x^2 + 1$ is a factor of the given polynomial, this polynomial may be written in factored form as
$x^4+x^3+2x^2+x+1 = (x^2+1) Q(x)$
$Q(x)$ is obtained by dividing numerator and denominator to obtain
$Q(x) = \dfrac{x^4+x^3+2x^2+x+1}{x^2+1} = x^2+x+1$
The given polynomial may be written as
$y = (x^2+1) (x^2+x+1 )$
The x-intercepts corresond to real zeros the polynomial that are obtained by solving
$(x^2+1) (x^2+x+1 ) = 0$
$x^2+1 = 0$ has no real solutions.
$x^2+x+1 = 0$ has no real solution since it is a quadratic equation wih discriminant $\Delta = (1)^2 - 4 \cdot 1 \cdot 1 = - 3$ negative.
The given polynomial has no real zeros and therefore no x-intercepts.

1. Introduction to Polynomials
2. Factor Polynomials
3. Real Zeros and Graphs of Polynomials
4. Polynomial Division
5. Solve Quadratic Equations Using Discriminants