# 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

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 $$4$$ .

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.

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.

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.