x^3-3x^2-x+3 factorise
Answers
Answer:
(
Answer:
(x - 3) (x + 1) (x - 1)
Step-by-step explanation:
(x + 1), (x - 1), and (x - 3) are the factors of x^3 - 3x^2 - x + 3x
3
−3x
2
−x+3
Step-by-step explanation:
Given that,
p(x) = x^3 - 3x^2 - x + 3x
3
−3x
2
−x+3
Using factor theorem,
The last term is 3 with factors 3 and 1.
so, by using trial and error
(x - 3) = 0
∵ x = 3
Thus,
p(3) = (3)^3 - 3(3)^2 - (3) + 3p(3)=(3)
3
−3(3)
2
−(3)+3
= 27 - 3 * 9
= 27 - 27
= 0
∵ (x - 3) is a factor of p(x)
by using trial and error
(x - 1) = 0
∵ x = 1
p(1) = (1)^3 - 1(1)^3 - 1 + 1p(1)=(1)
3
−1(1)
3
−1+1
= 1 - 1
= 0
∵ (x - 1) is a factor of p(x)
Thus,
p(x) = x^3 - 3x^2 - x + 3x
3
−3x
2
−x+3
= x^2 (x - 3) - 1 (x - 3)x
2
(x−3)−1(x−3)
= (x - 3) (x^2 - 1)(x−3)(x
2
−1)
= (x - 3) (x^2 - 1^2)(x−3)(x
2
−1
2
)
= (x - 3) [(x + 1) (x - 1)]=(x−3)[(x+1)(x−1)]
= = (x - 3) (x + 1) (x - 1)=(x−3)(x+1)(x−1)