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.
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Question 1Polynomial 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.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) = (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
Question 2The polynomial $$p(x)=3x^4+5x^3-17x^2-25x+10$$ 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) = (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
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 = (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.
Question 4Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $$p(x) = 6x^3-13x^2+x+2 $$.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. |
