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pochemuha
3 years ago
11

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

Mathematics
1 answer:
Anton [14]3 years ago
6 0

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

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