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

Question 1Polynomial p is defined by $$p(x) = x^3+5x^22x24 $$ has a zero at x = 2. Factor p completely and find its zeros.solutionp(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) = (x^{3} + 5 x^{2}  2 x  24) / (x  2) = x^{2} + 7 x + 12 Using the division above, p(x) may now be written in factored form as follows: p(x) = (x  2)(x^{2} + 7 x + 12) Factor the quadratic expression x^{2} + 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
Question 2The polynomial $$p(x)=3x^4+5x^317x^225x+10$$ has irrational zeros at x = ~+mn~ √5. Find the other zeros.solutionZeros 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) = (x^{2}  5)Q(x) Find Q(x) using long division of polynomials Q(x) = p(x) / (x^{2}  5) = (3 x^{4} + 5 x^{3}  17 x^{2}  25 x + 10) / (x^{2}  5) = 3 x^{2} + 5 x  2 Factor Q(x) = 3 x^{2} + 5 x  2 Q(x) = 3 x^{2} + 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
Question 3Polynomial 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. solutiona) 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} = (x^{4}  2x^{3}  2x^{2} + 6x  3) / (x  1)^{2} = x^{2}  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}(x^{2}  3) Factor the quadratic expression x^{2}  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. .
Question 4Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $$p(x) = 6x^313x^2+x+2 $$.solutionRational 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. 