# FINDING REAL ZEROS OF A POLYNOMIAL

**Definitions:**

• Polynomial: is a function of the form

The numbers
are called coefficients. a_{n} is the leading coefficient, a_{0} is the

constant term and the highest exponent n is the degree of the polynomial.

• Zero: If P is a polynomial and if c is a number such that P(c) = 0 then c is a zero of P .

• The following are all equivalent :

1. c is a zero of P.

2. x = c is an x-intercept of the graph of P .

3. x - c is a factor of P .

4. x = c is a solution of the equation P (x) = 0.

• Rational Zeros Theorem: If the polynomial

has integer coefficients, then every rational zero of P is of the form p/q where

p is a factor of the constant term a_{0};

and q is a factor of the leading coefficient a _{n}:

• The Rational Zeros Theorem does NOT list irrational
zeros. These will need to be found using other

means like the quadratic formula .

** Steps for finding the real zeros of a polynomial:**

1. List all possible rational zeros using the Rational
Zeros Theorem.

2. Use synthetic division to test the polynomial at each of the possible
rational zeros that you found in

step 1. (Remember that c is a zero when the remainder is zero.)

3. Repeat step 2 until you reach a quotient that is a quadratic or factors
easily. Use the quadratic

formula or factoring to find the remaining zeros.

** Common Mistakes to Avoid:**

• The Rational Zeros Theorem does NOT list the rational
zeros of P. It lists all POSSIBLE rational

zeros.

**PROBLEMS**

Find all real zeros of the polynomial.

1. P(x) = x^{3} - 7x^{2} + 14x - 8

Possible Zeros:

We start by trying 1 in synthetic division.

Remember that 1 is a zero if the remainder

is zero.

Therefore, x = 1 is a our first zero. Since the

quotient resulting from synthetic division is

always one degree less, the quotient that we

have is the quadratic x^{2} - 6x + 8 which can

easily be factored.

x^{2} - 6x + 8 = 0

(x - 2)(x - 4) = 0

Setting each factor equal to zero, we get

**Real zeros : 1; 2; 4**

2. P(x) = 2x^{3} + 15x^{2} + 22x - 15

Possible Zeros:

We start by trying -3 in synthetic division.

Remember that -3 is a zero if the remainder

is zero.

Therefore, x = -3 is a our first zero. Since

the quotient resulting from synthetic divi-

sion is always one degree less, the quotient

that we have is the quadratic 2x^{2} + 9x - 5

which can easily be factored.

2x^{2} + 9x - 5 = 0

(2x - 1)(x + 5) = 0Setting each factor equal to zero, we get

**Real zeros : **

3. P(x) = x^{3} - 22x - 15

Possible Zeros:

We start by trying 5 in synthetic division.

Remember that 5 is a zero if the remainder

is zero.

Therefore, x = 5 is a our first zero. Since the

quotient resulting from synthetic division is

always one degree less, the quotient that we

have is the quadratic x^{2} +5x+3 which can-

not be factored. Using the quadratic formula

to solve x^{2} + 5x + 3 = 0, we get

**Real zeros : **

4. P(x) = x^{3} + 3x^{2} + 6x + 4.

Possible Zeros:

We start by trying -1 in synthetic division.

Remember that -1 is a zero if the remainder

is zero.

Therefore, x = -1 is a our first zero. Since

the quotient resulting from synthetic divi-

sion is always one degree less, the quotient

that we have is the quadratic

x^{2} +2x+4 which cannot be factored. Using

the quadratic formula to solve x^{2}+2x+4 = 0,

we get

Because there is a negative under the square

root , these zeros are not real. Therefore, we

have no further real zeros.

**Real zeros : - 1**

5. P(x) = 2x^{4} + x^{3} - 16x^{2} + 3x + 18.

Possible Zeros:

We start by trying 2 in synthetic division.

Remember that 2 is a zero if the remainder

is zero.

Therefore, x = 2 is a our first zero. However,

our quotient is not a quadratic. Therefore,

we will try to find one more zero by using

synthetic division. We will try -1.

Hence, x = -1 is our second zero. Since the

quotient resulting from this synthetic divi-

sion is the quadratic 2x^{2} + 3x - 9, we will

factor this to solve.

2x^{2} + 3x - 9 = 0

(2x - 3)(x + 3) = 0

Setting each factor equal to zero, we get

**Real zeros : **

6. P(x) = 2x^{4} + 7x^{3} - x^{2} - 15x - 9

Possible Zeros:

We start by trying -1 in synthetic division.

Remember that -1 is a zero if the remainder

is zero.

Therefore, x = -1 is our first zero. However,

our quotient is not a quadratic. Therefore,

we will try to find one more zero by using

synthetic division. We will try -3.

Hence, x = -3 is our second zero. Since

the quotient resulting from this synthetic di-

vision is the quadratic 2x^{2} - x - 3, we will

factor this to solve.

2x^{2} - x - 3 = 0

(2x - 3)(x + 1) = 0

Setting each factor equal to zero, we get

**Real zeros :**

NOTE: x = -1 is a double zero. It occurs

two times . However, you only need to list it

once.

7. P(x) = 6x^{4} - 8x^{3} - 41x^{2} + 23x + 30.

Possible Zeros:

We start by trying 3 in synthetic division.

Remember that 3 is a zero if the remainder

is zero.

Therefore, x = 3 is a our first zero. However,

our quotient is not a quadratic. Therefore,

we will try to find one more zero by using

synthetic division. We will try

Hence, x = -2/3 is our second zero. Since the

quotient resulting from this synthetic divi-

sion is the quadratic 6x^{2} + 6x - 15, we need

to solve this.

6x^{2} + 6x - 15 = 0

3(2x^{2} + 2x - 5) = 0

Using the quadratic formula to solve

2x^{2} + 2x - 5 = 0, we get

**Real zeros : **

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