(x+2)(x+3)-(x-1)(x+4) solve
Answers
Answered by
0
Answer:
fghh
Step-by-step 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
i
Answered by
1
Answer:
(x+2)(x+3)- (x-1)(x+4)
x×x becomes 2x and 2×3 becomes6 -x×x becomes 2x and 1×4 becomes 4 so,
2x+6-2x+4
2x -2x becomes x (0) and 6-4 becomes 2 so,
x is 2
I DON'T KNOW IF IT'S THE CORRECT ANSWER
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