College Algebra Questions and Problems With Solutions
sample 6 : Polynomials

College algebra questions and problems on multiplying, dividing and finding polynomials are presented. The solutions are at the bottom of the page.

Question 1

Find the product of the polynomials P(x) = 2x 2 - 3x and Q(x) = 3x 2 + x - 5

Question 2

Find the quotient and the remainder when polynomial P(x) = 2x 4 - x 3 -3x 2 + 7x - 13 is divided by Q(x) = x 2 - 4

Question 3

Find the fourth degree polynomial P(x) whose graph is shown below.
college algebra problem 3, graph of fourth degree polynomial .
Figure 1. graph of fourth degree polynomial, question 3

Question 4

Find the remainder if 4 x 200 + 5 x 95 - 4 x 21 + 2x - 6 is divided by x - 1

Question 5

Find the third degree polynomial with leading coefficient -2 and having zeros 1, -2 and 3.

Question 7

Find all zeros of the polynomial P(x) = 3x 4 + 26 x 3 + 74 x 2 + 74 x + 15 knowing that P(x) has two rational zeros.

Question 7

Polynomial P(x) = x 3 + x 2 + c x + d has x - 2 as a factor and when divided by x - 1, the remainder is equal to 3. Find the constant c and d.

Question 8

P(x) is an even fourth degree polynomial function such that P(-2) = 0, P(x + 3) has a zero at x = 4 and P(0) = -9. Find P(x).

Question 9

The polynomial P(x) = 2 x 4 + 3 x 3 - 16 x 2 - 17 x + 12 has zeros at x = 1/2 and x = -3. Find the other zeros.

Question 10

Factor the following polynomials:
a) f(x) = x 3 - x 2 - 4 x + 4
b) f(x) = 2 x 2 - x 3
c) f(x) = x 2 - 2 x 4
d) f(x) = -(x 2 - 2 x - 3) 2
e) f(x) = x 4 + 3 x 3 + 3 x 2 + x

Question 11

A polynomial of degree 4 has a negative 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 below the x axis or above the x axis?

Question 12

Match the function to the graph.
1) f(x) = (x + 1)(x - 1) 2(x + 2) 2
2) f(x) = -(x + 1)(x - 1) 4
3) f(x) = (x + 1)(x - 1) 3(x - 3)
4) f(x) = (x + 1) 2(x - 1) 3
college algebra question 12, matching graph with functions
Figure 2. graph of 4 functions , question 12 matching

Question 13

Give four different reasons why the graph below cannot possibly be the graph of p(x) = x 4 - x 2 + 1.
college algebra question 13, graph of polynomial .
Figure 3. graph polynomial , question 13

Question 14

The graph of polynomial f is shown below.
a) Is the degree of f even or odd?
b) Is the leading coefficient negative or positive?
c) Can you find the degree of f from the graph?Explain your answer.
college algebra problem 14, graph of polynomial .
Figure 4. graph polynomial , question 14

Question 15

If the polynomial 3x3 + q x2 + p x + 7, where q and p are real numbers, is divided by x2 - 1, the remainder is 5x + 4. Find q and p.

Solutions to the Above Questions

Solution to Question 1

multiply polynomial question 1 solution.
Figure 5. polynomial multiplication , solution to question 1
P(x)Q(x) = (2x 2 - 3x)(3x 2 + x - 5)
= 6x 4 - 7x 3 -13x 2 + 15x

Solution to Question 2

divide polynomial question 2 solution.
Figure 6. polynomial division , solution to question 2
P(x)/ Q(x) = (2x 4 - x 3 -3x 2 + 7x - 13) / (x 2 - 4)
= 2x 2 - x + 5 + (3x + 7) / (x 2 - 4)
quotient = 2x 2 - x + 5 , remainder = 3x + 7

Solution to Question 3

Zeros at 0, 1, 3 and 4, hence P(x) can be written as P(x) = A x (x - 1)(x - 3)(x - 4)
We now use the point (2 , 2) to write P(2) = 2 and use it to find the constant A = 1/2.
P(x) = (1/2) x (x - 1)(x - 3)(x - 4)
Expand to find
P(x) = (1/2)x 4 - 4x 3 + (19/2)x 2 - 6x

Solution to Question 4

Using remainder theorem, remainder = P(1) = 4*1 200 + 5*1 95 - 4*1 21 + 2*1 - 6 = 1

Solution to Question 5

P(x) = -2(x - 1)(x + 2)(x - 3) = -2 x 3 + 4 x 2 + 10 x - 12

Solution to Question 6

According to the rational zeros theorem, the possible rational zeros are as follows: ~+mn~1 , ~+mn~1/3 , ~+mn~5/3 , ~+mn~3 , ~+mn~5
Substituting or using synthetic division, it can easily be shown that -3 and -5/3 are rational zeros of P(x). hence P(x) may be written as
P(x) = (x + 3)(x + 5/3) Q(x)
Q(x) = P(x) / (x + 3)(x + 5/3) = 3x 2 + 12 x + 3 (using division)
The remaining zeros of P(x) are the zeros of Q(x) which are -2 + √3 and -2 - √3.

Solution to Question 7

If (x - 2) is a factor, then P(2) = 0. The remainder theorem also states that P(1) = 3. After substitutions P(2) and P(1), we obtain a system of 2 equations in c and d.
2c + d = -12 and c + d = 1
solve to find: c = -13 and d = 14

Solution to Question 8

If P(-2) = 0, then -2 is a zero of P. Since P is even 2 is also a zero of P because of the symmetry of the graph of P with respect to y axis.
If P(x + 3) has a zero at x = 4, then P(x) has a zero at x = 1 since the graph of P(x + 3) is the graph of P(x) shifted 3 units left. Also and because of symmetry x = -1 is a zero of P. Hence P has 4 zeros: -2, -1, 1 and 2 and P(0) = -9.
P(x) = A (x + 1)(x + 2)(x - 1)(x - 2)
Use P(0) = -9 to find coefficient A = -9/4
P(x) = -(9/4) (x + 1)(x + 2)(x - 1)(x - 2) = -(9/4) x 4 + (45/4)x 2 - 9

Solution to Question 9

Since x = 1/2 and x = -3 are zeros of P, we can write
P(x) = (x - 1/2)(x + 3) Q(x)
Q(x) = P(x) / (x - 1/2)(x + 3) = 2(x 2 - x - 4) (using division)
The remaining zeros of P(x) are the zeros of Q(x): 1/2 + (1/2)√17 and 1/2 - (1/2)√17.

Solution to Question 10

a) f(x) = x 3 - x 2 - 4 x + 4
= x 2(x - 1) - 4(x - 1)
= (x - 1)(x 2 - 4)
= (x - 1)(x - 2)(x + 2)
b) f(x) = 2 x 2 - x 3
= x 2(2 - x)
c) f(x) = x 2 - 2 x 4
= x 2(1 - 2x 2)
= x 2(1 - x&radic2)(1 + x&radic2)
d) f(x) = -(x 2 - 2 x - 3) 2
= -((x + 1)(x - 3)) 2
= -(x + 1) 2(x - 3) 2
e) f(x) = x 4 + 3 x 3 + 3 x 2 + x
= x (x 3 + 3 x 2 + 3 x + 1)
= x(x + 1) 3

Solution to Question 11

Leading coefficient negative therefore between x = -2 and x = -1 graph is above x axis and below x axis between x = -1 and x = 1 and hence the y -intercept is below the x axis.

Solution to Question 12

1 - C
2 - D
3 - B
4 - A

Solution to Question 13

1) P is even , the graph is not symmetric with respect to y axis.
2) P has no zeros, graph has x intercepts.
3) P(0) = 1 , y intercept in graph below x axis.
4) Leading coefficient in P is negative, graph going down on the left and on the right.

Solution to Question 14

a) The degree of f odd since its graph has 5 x-intercepts and the complex zeros come in pairs.
b) The leading coefficient is negative because the graph is going down on the right and up on the left.
c) No, the degree of a polynomial is determined by real and complex zeros. Only the real zeros in the form of x-intercepts are shown on the graph.

Solution to Question 15

The given polynomial, the divisor (x2 - 1), the remainder 5x + 4 and the quotient Q(x) are related as follows:
3x3 + qx2 + px + 7 = Q(x)(x2 - 1) + 5x + 4.
Substitute x by 1 and -1 on both sides of the equation to obtain 2 equations in q and p.
x = 1: 3 + q + p + 7 = 0 + 9 (1)
x = -1: -3 + q - p + 7 = -1
Solve the above system of equations to find: q = -3 and p = 2.

More References and links

Introduction to Polynomials
Graphs of Polynomials Functions
Worksheet on Polynomials
Algebra Questions and problems
More ACT, SAT and Compass practice

Popular Pages

More Info