Find Zeros of Polynomials

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.

Questions with Solutions

Question 1

Polynomial

p

is defined by

p (x) = x^{3} + 5 x^{2} - 2 x - 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)

x - 2

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

has irrational zeros at

x = \pm \sqrt{5}

. Find the other zeros.

solution

Zeros at

x = \pm \sqrt{5}

correspond to the factors.

x - \sqrt{5}

and

x + \sqrt{5}

Hence polynomial

p (x)

may be written as

p (x) = (x - \sqrt{5}) (x + \sqrt{5}) Q (x) = (x^{2} - 5) Q (x)

Find

Q (x)

using long division of polynomials

Q (x) = \frac{p (x)}{x^{2} - 5}

= \frac{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 = (3 x - 1) (x + 2)

Factor

p (x)

completely

p (x) = (x - \sqrt{5}) (x + \sqrt{5}) (3 x - 1) (x + 2)

Set each of the factors of p(x) to zero to find the zeros.

x = \pm \sqrt{5}, x = \frac{1}{3}, x = - 2

Question 3

Polynomial

p

is given by

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

(x - 1)^{2}

must give a remainder equal to

0

. A long division gives

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

The remainder in the division of

p (x)

(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) = 6 x^{3} - 13 x^{2} + x + 2

solution

Rational Zeros Theorem: If

p (x)

is a polynomial with integer coefficients and if

\frac{m}{n}

(in lower terms) is a zero of

p (x)

, then

m

is a factor of the constant term

2

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 :

\pm 1

\pm 2

factors of

6

are:

\pm 1, \pm 2, \pm 3, \pm 6

possible zeros: divide factors of

2

by factors of

6

\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{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

- \frac{1}{3}

\frac{1}{2}

and

2

polynomials question 4 .

We now calculate

p (- \frac{1}{3}), p (\frac{1}{2})

and

p (2)

to finally check if these are the exact zeros of

p (x)

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

p (\frac{1}{2}) = 6 {(\frac{1}{2})}^{3} - 13 {(\frac{1}{2})}^{2} + (\frac{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

- \frac{1}{3}, \frac{1}{2}

and

2

Find Zeros of Polynomials

Questions with Solutions

Question 1

solution

Question 2

solution

Question 3

solution

Question 4

solution

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