What are the zeros of the polynomial function? f (x) = x^4-4x^3-22x^2+4x+21

-3

-1

0

1

3

7

Replace

f

(

x

)

f

(

x

)

with

y

y

.

y

=

2

x

3

-

4

x

2

-

22

x

+

44

y

=

2

x

3

-

4

x

2

-

22

x

+

44

To find the roots of the equation, replace

y

y

with

0

0

and solve.

0

=

2

x

3

-

4

x

2

-

22

x

+

44

0

=

2

x

3

-

4

x

2

-

22

x

+

44

Rewrite the equation as

2

x

3

-

4

x

2

-

22

x

+

44

=

0

2

x

3

-

4

x

2

-

22

x

+

44

=

0

.

2

x

3

-

4

x

2

-

22

x

+

44

=

0

2

x

3

-

4

x

2

-

22

x

+

44

=

0

Factor the left side of the equation.

Tap for more steps ...

2

(

x

-

2

)

(

x

2

-

11

)

=

0

2

(

x

-

2

)

(

x

2

-

11

)

=

0

Divide both sides of the equation by

2

2

. Dividing

0

0

by any non-zero number is

0

0

.

(

x

-

2

)

(

x

2

-

11

)

=

0

(

x

-

2

)

(

x

2

-

11

)

=

0

Set

x

-

2

x

-

2

equal to

0

0

and solve for

x

x

.

Tap for more steps ...

x

=

2

x

=

2

Set

x

2

-

11

x

2

-

11

equal to

0

0

and solve for

x

x

.

Tap for more steps ...

x

=

√

11

,

-

√

11

x

=

11

,

-

11

The solution is the result of

x

-

2

=

0

x

-

2

=

0

and

x

2

-

11

=

0

x

2

-

11

=

0

.

x

=

2

,

√

11

,

-

√

11