Fourth Degree Polynomials

This page presents several fourth degree polynomials along with questions and detailed solutions. Topics include graphs, x-intercepts, multiplicities, and parameter determination.

Questions

Question 1

About: Polynomial of degree four that touches the x-axis at one point.

Question: Why does the graph touch (but not cross) the x-axis at only one point?

Graph of y = x^4 — Fig.1 - Graph of the Fourth Polynomial \( y = x^4 \)

Question 2

About: Polynomial of degree four with two x-intercepts.

Question: If the graph cuts the x-axis at \(x = 1\), what is the coordinate of the other x-intercept?

Graph of a fourth degree polynomial with 2 x-intercepts — Fig.2 - Graph of the Fourth Polynomial \( y = x^4+0.5x-x^3-0.5 \)

Question 3

About: Polynomial of degree four with three x-intercepts and a parameter \(a\) to determine.

Question: The graph touches (but does not cross) the x-axis at \(x = 2\). What are the coordinates of the other two x-intercepts?

Graph of a fourth degree polynomial with 3 x-intercepts — Fig.3 - Graph of the Fourth Polynomial \( y = x^4-2x^3-5x^2+ a x-4 \)

Question 4

About: Polynomial of degree four with no x-intercepts.

Question: Why does the graph of \( y = x^4+x^3+2x^2+x+1 \) have no x-intercept, knowing that \(x^2+1\) is a factor of this polynomial?

Graph of a fourth degree polynomial with no x-intercepts — Fig.4 - Graph of the Fourth Polynomial \( y = x^4+x^3+2x^2+x+1 \)

Solutions

Answer Question 1

Polynomial: \( y = x^4 \). Solving \( x^4 = 0 \) gives a zero of multiplicity 4. Therefore, the graph touches the x-axis at one point and is flat at \(x = 0\) indicating multiplicity 4.

Answer Question 2

Given an x-intercept at \( x = 1 \), \( x-1 \) is a factor. The polynomial can be written as: \[ y = x^4+0.5x-x^3-0.5 = (x-1)Q(x) \] Using polynomial division: \[ Q(x) = \frac{x^4+0.5x-x^3-0.5}{x-1} = x^3+0.5 \] The other zero is obtained by solving \( x^3 + 0.5 = 0 \): \[ x = -\sqrt[3]{0.5} \approx -0.8 \]

Answer Question 3

The graph touches the x-axis at \( x = 2 \). Solve for \(a\) in: \[ 2^4-2(2)^3-5(2)^2+ a(2)-4 = 0 \Rightarrow a = 12 \] Polynomial becomes: \[ y = x^4-2x^3-5x^2+12x-4 \] Since \( x = 2 \) has even multiplicity (2) and total degree is 4, we factor: \[ y = (x-2)^2 Q(x)\] Using polynomial division: \[ Q(x) = x^2+2x-1 \] Solve for remaining zeros: \[ x^2+2x-1=0 \Rightarrow x=-1+\sqrt{2} \approx 0.41,\quad x=-1-\sqrt{2} \approx -2.41 \]

Answer Question 4

Factorization: \[ x^4+x^3+2x^2+x+1 = (x^2+1)Q(x),\quad Q(x) = x^2+x+1 \] Solving \(Q(x)=0\) gives discriminant: \[ \Delta = 1^2 - 4(1)(1) = -3 < 0 \] No real solutions exist, so the polynomial has no real zeros and no x-intercepts.