Factorize (x-1)(x + 2)(x-3)(x+4) + 24
Answers
Answer:
x
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
ix
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
ix
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
ix
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
ix
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
ix
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
0
5
5
2
±
√
15
2
i
Explanation:
Let
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
Then:
f
(
0
)
=
(
−
1
)
(
−
2
)
(
−
3
)
(
−
4
)
=
4
!
=
24
f
(
5
)
=
(
5
−
1
)
(
5
−
2
)
(
5
−
3
)
(
5
−
4
)
=
4
⋅
3
⋅
2
⋅
1
=
4
!
=
24
So both
x
=
0
and
x
=
5
are roots and
x
and
(
x
−
5
)
are factors.
f
(
x
)
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
−
24
=
x
4
−
10
x
3
+
35
x
2
−
50
x
=
x
(
x
−
5
)
(
x
2
−
5
x
+
10
)
The remaining quadratic factor is in the form
a
x
2
+
b
x
+
c
, with
a
=
1
,
b
=
−
5
and
c
=
10
.
This has zeros given by the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
5
±
√
5
2
−
(
4
⋅
1
⋅
10
)
2
=
5
±
√
−
15
2
=
5
2
±
√
15
2
Step-by-step explanation: