Find Zeros of Polynomials

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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 1

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 \( x - 2 \)

\( \dfrac{p(x)}{x - 2} = \dfrac {x^3 + 5 x^2 - 2 x - 24}{x-2} = x^2 + 7 x + 12 \)

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 2

The polynomial \( p(x)=3x^4+5x^3-17x^2-25x+10 \) has irrational zeros at \( x = \pm \sqrt5 \). Find the other zeros.

solution

Zeros at \( x = \pm \sqrt5 \) correspond to the factors.
\( x - \sqrt 5 \) and \( x + \sqrt 5 \)
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
\( Q(x) = \dfrac{p(x)}{x^2 - 5} \)

\( \quad = \dfrac{3 x^4 + 5 x^3 - 17 x^2 - 25 x + 10}{x^2 - 5} \)

\( \quad = 3 x^2 + 5 x - 2 \)
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
\( p(x) = (x - \sqrt 5)(x + \sqrt 5)(3x - 1)(x + 2) \)
Set each of the factors of p(x) to zero to find the zeros.
\( x = \pm \sqrt 5 , x = \dfrac{1}{3} , x = - 2 \)

Question 3

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

\( \dfrac{p(x)}{(x - 1)^2} = \dfrac{x^4 - 2x^3 - 2x^2 + 6x - 3}{(x - 1)^2} = x^2 - 3 \)

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.
\( p(x) = (x - 1)^2 (x - \sqrt 3) (x + \sqrt 3) \)
The zeros are found by solving the equation.
\( p(x) = (x - 1)^2 (x - \sqrt 3) (x + \sqrt 3) = 0 \)
For \( p(x) \) to be equal to zero, we need to have
\( (x - 1)^2 = 0 \) , or \( x - \sqrt 3 = 0 \) , or \( x + \sqrt 3 = 0 \)
Solve each of the above equations to obtain the zeros of p(x).
\( x = 1 \) (multiplicity \( 2 \) ) , \( x = \sqrt 3 \) and \( x = - \sqrt 3 \)
c) With the help of the factored form of \( p(x) \) and its zeros found above, we now make a table of signs.

table of sign question 3 .

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.

polynomials question 1 .

Question 4

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 \( \dfrac{m}{n} \) (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 \( 2 \) are : \( \quad \pm 1 \) , \( \pm 2 \)
factors of \( 6 \) are: \( \quad \pm 1 , \pm 2 , \pm 3 , \pm 6 \)
possible zeros: divide factors of \( 2 \) by factors of \( 6 \): \( \quad \pm 1 , \pm \dfrac{1}{2} , \pm \dfrac{1}{3} , \pm \dfrac{1}{6} , \pm 2 , \pm \dfrac{2}{3} \)
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 \( - \dfrac{1}{3} \), \( \dfrac{1}{2} \) and \( 2 \).

polynomials question 4 .

We now calculate \( p(-\dfrac{1}{3}), p(\dfrac{1}{2}) \) and \( p(2) \) to finally check if these are the exact zeros of \( p(x) \).

\( p \left(-\dfrac{1}{3} \right) = 6\left(-\dfrac{1}{3} \right)^3 - 13 \left (-\dfrac{1}{3} \right)^2 + \left (-\dfrac{1}{3} \right) + 2 = 0 \)

\( p \left(\dfrac{1}{2} \right) = 6 \left (\dfrac{1}{2} \right)^3 - 13 \left (\dfrac{1}{2} \right)^2 + \left (\dfrac{1}{2} \right) + 2 = 0 \)

\( p(2) = 6(2)^3 - 13(2)^2 + (2) + 2 = 0 \)
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 \( -\dfrac{1}{3}, \dfrac{1}{2} \) and \( 2 \).

Free Mathematics Tutorials

Find Zeros of Polynomials

Question 1

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Question 2

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Question 3

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Question 4

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