What are the zeros of the polynomial function? f(x)=x^4−4x^3−22x^2+4x+21
1 answer:
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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Step-by-step explanation:
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4n + 1 = 25
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Then divide by 4 on both sides:
4n = 24
n = 6
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