# Grade 12 How to Find Zeros of Polynomials Questions with Detailed Solutions

How to find the zeros of polynomials using factoring, division of polynomials and the rational zeros theorem. Grade 12 math questions are presented along with detailed solutions and graphical interpretations.

 Polynomial p is defined by $p(x) = x^3+5x^2-2x-24$ has a zero at x = 2. Factor p completely and find its zeros. Solution p(x) has a zero at x = 2 and therefore x - 2 is a factor of p(x). Divide p(x) by x - 2 p(x) / (x - 2) = (x3 + 5 x2 - 2 x - 24) / (x - 2) = x2 + 7 x + 12 Using the division above, p(x) may now be written in factored form as follows: p(x) = (x - 2)(x2 + 7 x + 12) Factor the quadratic expression x2 + 7 x + 12. p(x) = (x - 2)(x + 3)(x + 4) The zeros are found by solving the equation. p(x) = (x - 2)(x + 3)(x + 4) = 0 For p(x) to be equal to zero, we need to have x - 2 = 0 , or x + 3 = 0 , or x + 4 = 0 Solve each of the above equations to obtain the zeros of p(x). x = 2 , x = - 3 and x = - 4 The polynomial $p(x)=3x^4+5x^3-17x^2-25x+10$ has irrational zeros at x = ~+mn~ √5. Find the other zeros. Solution Zeros at x = ~+mn~ √5, correspond to the factors. (x - √5) and (x + √5) Hence polynomial p(x) may be written as p(x) = (x - √5)(x + √5) Q(x) = (x2 - 5)Q(x) Find Q(x) using long division of polynomials Q(x) = p(x) / (x2 - 5) = (3 x4 + 5 x3 - 17 x2 - 25 x + 10) / (x2 - 5) = 3 x2 + 5 x - 2 Factor   Q(x) = 3 x2 + 5 x - 2 Q(x) = 3 x2 + 5 x - 2 = (3x - 1)(x + 2) Factor p(x) completely p(x) = (x - √5)(x + √5)(3x - 1)(x + 2) Set each of the factors of p(x) to zero to find the zeros. x = ~+mn~√ 5 , x = 1 / 3 , x = - 2 Polynomial p is given by $p(x) = x^4 - 2x^3 - 2x^2 + 6x - 3$ a) Show that x = 1 is a zero of multiplicity 2. b) Find all zeros of p. c) Sketch a possible graph for p. Solution a) If x = 1 is a zero of multiplicity 2, then (x - 1)2 is a factor of p(x) and a division of p(x) by (x - 1)2 must give a remainder equal to 0. A long division gives p(x) / (x - 1)2 = (x4 - 2x3 - 2x2 + 6x - 3) / (x - 1)2 = x2 - 3 The remainder in the division of p(x) by (x - 1)2 is equal to 0 and therefore x = 1 is a zero of multiplicity 2. b) Using the division above, p(x) may now be written in factored form as follows p(x) = (x - 1)2(x2 - 3) Factor the quadratic expression x2 - 3. p(x) = (x - 1)2 (x - √3) (x + √3) The zeros are found by solving the equation. p(x) = (x - 1)2 (x - √3) (x + √3) = 0 For p(x) to be equal to zero, we need to have (x - 1)2 = 0 , or (x - √3) = 0 , or (x + √3) = 0 Solve each of the above equations to obtain the zeros of p(x). x = 1 (multiplicity 2) , x = √3 and x = - √3 c) With the help of the factored form of p(x) and its zeros found above, we now make a table of signs. . We use the zeros of p(x) which graphically are shown as x intercepts, the table of signs and the y intercept (0 , -3) to complete the graph as shown below. . Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $p(x) = 6x^3-13x^2+x+2$. Solution Rational Zeros Theorem: If p(x) is a polynomial with integer coefficients and if m / n (in lower terms) is a zero of p(x), then m is a factor of the constant term 2 of p(x) and n is a factor of the leading 6 coefficient of p(x). Find factors of 2 and 6. factors of 2: ~+mn~ 1 , ~+mn~ 2 factors of 6: ~+mn~ 1 , ~+mn~ 2 , ~+mn~ 3 , ~+mn~ 6 possible zeros: divide factors of 2 by factors of 6: ~+mn~ 1 , ~+mn~ 1 / 2 , ~+mn~ 1 / 3 , ~+mn~ 1 / 6 , ~+mn~ 2 , ~+mn~ 2 / 3 Because of the large list of possible zeros, we graph the polynomial and guess the zeros from the location of the x intercepts. Below is the graph of the the given polynomial p(x) and we can easily see that the zeros are close to -1/3, 1/2 and 2. . We now calculate p(-1/3), p(1/2) and p(2) to finally check if these are the exact zeros of p(x). p(-1/3) = 6(-1/3)^3 - 13(-1/3)^2 + (-1/3) + 2 = 0 p(1/2) = 6(1/2)^3 - 13(1/2)^2 + (1/2) + 2 = 0 p(2) = 6(2)^3 - 13(2)^2 + (2) + 2 = 0 We have used the rational zeros theorem and the graph of the given polynomial to determine the 3 zeros of the given polynomial which are -1/3, 1/2 and 2.

More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers

More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers

More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers

Author - e-mail