factorise x^3-2x^2-x+2
Answers
Answer:
Explanation:
x
3
−
2
x
2
−
x
+
2
=
x
3
−
x
−
2
x
2
+
2
Grouping the
1
s
t
2
terms together and the
2
n
d
2
together:
=
x
(
x
2
−
1
)
−
2
(
x
2
−
1
)
=
(
x
2
−
1
)
(
x
−
2
)
Using the identity:
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
=
(
x
2
−
1
2
)
(
x
−
2
)
=
(
x
−
1
)
(
x
+
1
)
(
x
−
2
)
Alternate Method:-
The above method is an easy on to solve this question. For factorizing other cubic polynomials, the following method can be used:
First, by trial and error method, you can find one factor as follows:
x
3
−
2
x
2
−
x
+
2
When replacing 1,
⇒
1
3
−
2
×
1
2
−
1
+
2
⇒
1
−
2
−
1
+
2
⇒
0
So, we get
(
x
−
1
)
as factor.
Then by long division, divide
(
x
−
1
)
by
(
x
3
−
2
x
2
−
x
+
2
)
You get
⇒
(
x
−
1
)
(
x
2
−
x
−
2
)
Then you have to factorize it by splitting the middle term method.
⇒
(
x
−
1
)
[
x
2
+
x
−
2
x
−
2
]
⇒
(
x
−
1
)
[
x
(
x
+
1
)
−
2
(
x
+
1
)
]
⇒
(
x
−
1
)
(
x
+
1
)
(
x
−
2
)
~Hope this helps! :)
Answer:
(x-1)(x+1)(x-2)
Step-by-step explanation:
Solution (i) Let take f(x) = x3 - 2x2 - x + 2
The constant term in f(x) is are ±1 and ±2
Putting x = 1 in f(x), we have
f(1) = (1)3 - 2(1)2 -1 + 2
= 1 - 2 - 1 + 2 = 0
According to remainder theorem f(1) = 0 so that (x - 1) is a factor of x3 - 2x2 - x + 2
Putting x = - 1 in f(x), we have
f(-1) = (-1)3 - 2(-1)2 –(-1) + 2
= -1 - 2 + 1 + 2 = 0
According to remainder theorem f(-1) = 0 so that (x + 1) is a factor of x3 - 2x2 - x + 2
Putting x = 2 in f(x), we have
f(2) = (2)3 - 2(2)2 –(2) + 2
= 8 -82 - 2 + 2 = 0
According to remainder theorem f(2) = 0 so that (x – 2 ) is a factor of x3 - 2x2 - x + 2
Here maximum power of x is 3 so that its can have maximum 3 factors
So our answer is (x-1)(x+1)(x-2)