The remaining zeros may found by solving the equation
3 x 2 + 5 x - 2 = 0
Solve the above equation to find the remaining zeros of f.
-2 and 1 / 3
A polynomial of degree 4 has a positive leading coefficient and simple zeros (i.e. zeros of multiplicity 1) at x = 2, x = - 2, x = 1 and x = -1. Is the y intercept of the graph of this polynomial positive or negative?
Solution to Problem 5:
All polynomials with degree 4 and positive leading coefficient will have a graph that rises to the left and to the right. And since the polynomial has two negative zeros and two positive zeros, then the only possibility for the y intercept is to be positive.
The graph of polynomial p is shown below.
a - Is the degree of p even or odd?
b - Is the leading coefficient negative or positive?
c - Can you find the degree from the graph of p? Explain.
Solution to Problem 6:
a - odd
b - negative
c - No. The degree of a polynomial depends on the real and complex zeros. The graph shows only the real zeros. Hence, not enough information is given to find the degree of the polynomial.
Give 4 different reasons why the graph below cannot be the graph of the polynomial p give by.
p(x) = x 4 - x 2 + 1
Solution to Problem 7:
1 - The given polynomial has degree 4 and positive leading coefficient and the graph should rise on the left and right sides.
2 - p(0) = 1, graph shows negative value.
3 - the equation x 4 - x 2 + 1 = 0 has no solution which suggests that the polynomial p(x) = x 4 - x 2 + 1 has no zeros. The graph shows x intercepts.
4 - The graph has a zero of multiplicity 2, a zero of multiplicity 3 and and a zero of multiplicity 1. So the degree of the graphed polynomial should be at least 6. The given polynomial has degree 4.