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 1Find the product of the polynomials P(x) = 2x^{ 2}  3x and Q(x) = 3x^{ 2} + x  5Question 2Find the quotient and the remainder when polynomial P(x) = 2x^{ 4}  x^{ 3} 3x^{ 2} + 7x  13 is divided by Q(x) = x^{ 2}  4Question 3Find the fourth degree polynomial P(x) whose graph is shown below.Question 4Find the remainder if 4 x^{ 200} + 5 x^{ 95}  4 x^{ 21} + 2x  6 is divided by x  1Question 5Find the third degree polynomial with leading coefficient 2 and having zeros 1, 2 and 3.Question 7Find 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 7Polynomial 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 8P(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 9The 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 10Factor 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 11A 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 12Match 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} Question 13Give four different reasons why the graph below cannot possibly be the graph of p(x) = x^{ 4}  x^{ 2} + 1.Question 14The 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. Question 15If the polynomial 3x^{3} + q x^{2} + p x + 7, where q and p are real numbers, is divided by x^{2}  1, the remainder is 5x + 4. Find q and p.
Solutions to the Above QuestionsSolution to Question 1P(x)Q(x) = (2x^{ 2}  3x)(3x^{ 2} + x  5)= 6x^{ 4}  7x^{ 3} 13x^{ 2} + 15x Solution to Question 2P(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 3Zeros 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 4Using remainder theorem, remainder = P(1) = 4*1^{ 200} + 5*1^{ 95}  4*1^{ 21} + 2*1  6 = 1Solution to Question 5P(x) = 2(x  1)(x + 2)(x  3) = 2 x^{ 3} + 4 x^{ 2} + 10 x  12Solution to Question 6According to the rational zeros theorem, the possible rational zeros are as follows: ~+mn~1 , ~+mn~1/3 , ~+mn~5/3 , ~+mn~3 , ~+mn~5Substituting 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 7If (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 8If 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 9Since x = 1/2 and x = 3 are zeros of P, we can writeP(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 10a) 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 11Leading 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 121  C2  D 3  B 4  A Solution to Question 131) 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 14a) The degree of f odd since its graph has 5 xintercepts 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 xintercepts are shown on the graph. Solution to Question 15The given polynomial, the divisor (x^{2}  1), the remainder 5x + 4 and the quotient Q(x) are related as follows:3x^{3} + qx^{2} + px + 7 = Q(x)(x^{2}  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 linksIntroduction to PolynomialsGraphs of Polynomials Functions Worksheet on Polynomials Algebra Questions and problems More ACT, SAT and Compass practice
