Polynomial Problems with Solutions

Polynomial problems with detailed solutions.



Problem 1:

The graph of a cubic polynomial

y = a x 3 + b x 2 + c x + d


is shown below. Find the coefficients a, b, c and d.

graph of polynomial, problem 1..

Solution to Problem 1:

  • This polynomial has a zero of multiplicity 1 at x = -2 and a zero of multiplicity 2 at x = 1. Hence the polynomial may be written as

    y = a (x + 2)(x - 1) 2

  • This polynomial has a y intercept (0 , 1). Hence

    1 = a (0 + 2)(0 - 1) 2

  • Solve for a to obtain

    a = 1 / 2

  • The polynomial may now be written as follows

    y = (1 / 2) (x + 2)(x - 1) 2

  • Expand to obtain

    y = (1 / 2) x 3 -(3 / 2) x + 1

  • We now identify the coefficients as follows

    a = 1/2 , b = 0, c = -3/2 and d = 1




Problem 2:

The graph of the polynomial

y = a x 4 + b x 3 + c x 2 + d x + e


is shown below. Find the coefficients a, b, c, d and e.

graph of polynomial, problem 2..

Solution to Problem 2:

  • This polynomial has a zero of multiplicity 2 at x = - 2 and a zero of multiplicity 2 at x = 2. Hence it may be written as

    y = a (x + 2) 2 (x - 2) 2

  • We now use the y intercept at (0 , -2) to write the equation

    -2 = a (0 + 2) 2 (0 - 2) 2

  • Solve the above for a to obtain

    a = - 1 / 8

  • We now write the polynomial as follows

    y = (-1 / 8) (x + 2) 2 (x - 2) 2

  • Expand

    y = (-1 / 8) x 4 + x 2 - 2

  • We now identify the coefficients

    a = -1 / 8, b = 0, c = 1, d = 0, e = 1


Problem 3:

The polynomial

f(x) = x 6 + 4 x 5 + x 4 - 12 x 3 - 11 x 2 + 4 x + 4


has a zero of multiplicity 2 at x = - 2. Find the other real zeros.

Solution to Problem 3:

  • If f has a zero of multiplicity 2, then it may be written as follows

    f(x) = (x + 2) 2 Q(x)

  • Where Q(x) is a polynomial of degree 4 and may be found by division

    Q(x) = f(x) / (x + 2) 2 = x 4 -3 x 2 + 1

  • Polynomial f may now be written as

    f(x) = (x + 2) 2 (x 4 -3 x 2 + 1)

  • The remaining zeros of polynomial f may be found by solving the equation

    x 4 -3 x 2 + 1 = 0

  • It is an equation of the quadratic type with solutions

    ( sqrt(5) + 1 ) / 2 , ( sqrt(5) - 1 ) / 2 , ( - sqrt(5) - 1 ) / 2 , ( - sqrt(5) + 1 ) / 2


Problem 4:

The polynomial

f(x) = 3 x 4 + 5 x 3 - 17 x 2 - 25 x + 10


has irrational zeros at + sqrt(5) and - sqrt(5). Find the other zeros.

Solution to Problem 4:

  • polynomial f may be written as

    f(x) = (x + sqrt(5)) (x - sqrt(5)) Q(x) = (x 2 - 5) Q(x)

  • Q(x) may be found by division

    Q(x) = f(x) / (x 2 - 5) = 3 x 2 + 5 x - 2

  • Hence f(x) may be written as

    f(x) = (x 2 - 5) (3 x 2 + 5 x - 2 )

  • 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


Problem 5:

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.


Problem 6:

The graph of polynomial p is shown below.

graph of polynomial, problem 6..

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.


Problem 7:

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


graph of polynomial, problem 7..

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.

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Updated: 2 April 2013

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