Polynomial Functions, Zeros, Factors and Intercepts (1)
Tutorial and problems with detailed solutions on finding polynomial functions given their zeros and/or graphs and other information. Problems related to polynomials with real coefficients and complex solutions are also included.
ReviewA) Let p(x) be a polynomial function with real coefficients. If p(s) = 0
B) In what follows the imaginary unit i is defined as i = √(-1) Let p(x) be a polynomial function with real coefficients. If a + ib is an complex zero of p(x), the conjugate a - bi is also a zero of p(x). Examplelet p(x) = x3 - 2 x2 + 2 x - 4
 Problems with Solutions
Problem 1The graph below is that of a polynomial function p(x) with real coefficients. The degree of p(x) is 3 and the zeros are assumed to be integers. Find p(x). The graph has 2 x intercepts: -1 and 2. The x intercept at -1 is of multiplicity 2. p(x) can be written as follows p(x) = a(x + 1)2(x - 2) , a is any real constant not equal to zero. To find a we need to use more information in the graph. The y intercept is at (0 , -2), which means that p(0) = -2 a(0 + 1)2(0 - 2) = -2 Solve the above equation for a to obtain a = 1 p(x) is given by p(x) = (x + 1)2(x - 2)
Problem 2A polynomial function p(x) with real coefficients and of degree 5 has the zeros: -1, 2(with multiplicity 2) , 0 and 1. p(3) = -12. Find p(x).Solution to Problem 2 p(x) can be written as follows p(x) = a x(x + 1)(x - 2)2(x - 1) , a is any real constant not equal to zero. p(3) = -12 gives the following equation in a. a(3)(3 + 1)(3 - 2)2(3 - 1) = -12 Solve the above equation for a to obtain a = -1/2 p(x) is given by p(x) = -0.5 x(x + 1)(x - 2)2(x - 1) The graph of p(x) is shown below. Check the intercepts and the point (3 , -12) on the graph of p(x) found above.
Problem 32 + i is a zero of polynomial p(x) given below, find all the other zeros.Solution to Problem 3 The zero 2 + i is a complex number and p(x) has real coefficients. It follows that the conjugate 2 - i is also a zero of p(x). p(x) may be written in factored form as follows p(x) = (x - (2 + i)) (x - (2 - i)) q(x) Let us expand the term (x - (2 + i)) (x - (2 - i)) in p(x) (x - (2 + i)) (x - (2 - i)) = x2 -(2 + i)x -(2 - i)x + (2+i)(2-i) = x2 - 4 x + 5 q(x) can be found by dividing p(x) by x2 - 4 x + 5. (x4 - 2 x3 - 6 x2 + 22 x - 15) / (x2 - 4 x + 5) = x2 + 2 x - 3 We now write p(x) in factored form p(x) = (x - (2 + i)) (x - (2 - i)) (x2 + 2 x - 3) The remaining 2 zeros of p(x) are the solutions to the quadratic equation. x2 + 2 x - 3 = 0 Factor the above quadratic equation and solve. (x - 1) (x + 3) = 0 solutions x = 1 x = -3 p(x) has the following zeros. 2 + i , 2 - i, -3 and 1.
Problem 4-3 - i is a zero of polynomial p(x) given below, find all the other zeros.Solution to Problem 4 The zero -3 - i is a complex number and p(x) has real coefficients. Hence the conjugate -3 + i is also a zero of p(x). The factored form of p(x) is as follows p(x) = (x - (-3 - i)) (x - (-3 + i)) q(x) Let us expand the term (x - (-3 - i)) (x - (-3 + i)) in p(x) (x - (-3 - i)) (x - (-3 + i)) = x2 + 6 x + 10 q(x) can be found by dividing p(x) by x2 + 6 x + 10. (x4 + 6 x3 + 11 x2 + 6 x + 10) / (x2 + 6 x + 10) = x2 + 1 We now write p(x) in factored form p(x) = (x - (2 + i)) (x - (2 - i)) (x2 + 1) The remaining 2 zeros of p(x) are the solutions to the quadratic equation. x2 + 1 = 0 Factor the above quadratic equation and solve. x2 + 1 = (x - i)(x + i) solutions x = i x = - i p(x) has the following zeros. - 3 - i , - 3 + i, i and - i. More References and Links to Polynomial FunctionsFactor Polynomials.Polynomial Functions. Multiplicity of Zeros and Graphs Polynomials. Find Zeros of Polynomial Functions - Problems Graphs of Polynomial Functions - Self Test. Step by Step Solver to Find a Polynomial Given its Zeros and a Point. Step by Step Solver to Factor a Cubic Polynomial Given one of its Zeros. Find a Cubic Polynomial Passing Through Four Points. |
