factorise by the division of polynomials: x^3-5x^2+2x+24÷x-4
Answers
Answered by
13
Answer:
Let
f
(
x
)
=
x
3
−
5
x
2
−
2
x
+
24
By the rational root theorem, any rational zeros of
f
(
x
)
must be expressible in the for
p
q
for integers
p
,
q
with
p
a divisor of the constant term
24
and
q
a divisor of the coefficient
1
of the leading term.
That means that the only possible rational zeros are the factors of
24
, namely:
±
1
,
±
2
,
±
3
,
±
4
,
±
6
,
±
12
,
±
24
Try each in turn:
f
(
1
)
=
1
−
5
−
2
+
24
=
18
f
(
−
1
)
=
−
1
−
5
+
2
+
24
=
20
f
(
2
)
=
8
−
20
−
4
+
24
=
8
f
(
−
2
)
=
−
8
−
20
+
4
+
24
=
0
So
x
=
−
2
is a zero and
(
x
+
2
)
is a factor.
x
3
−
5
x
2
−
2
x
+
24
=
(
x
+
2
)
(
x
2
−
7
x
+
12
)
We can factor
x
2
−
7
x
+
12
by noting that
4
×
3
=
12
and
4
+
3
=
7
, so:
x
2
−
7
x
+
12
=
(
x
−
4
)
(
x
−
3
)
Putting it all together:
x
3
−
5
x
2
−
2
x
+
24
=
(
x
+
2
)
(
x
−
4
)
(
x
−
3
)
Answered by
0
Answer:
Step-by-step explanation:
By long division,
with a remainder of 16.
Now
∴
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