Question

Factorise by using factor theorem x3 - 3x2 - x + 3

Answer 1

Answer:

Step-by-step explanation:

Answer 2

(x + 1), (x - 1), and (x - 3) are the factors of $x^3 - 3x^2 - x + 3$

Step-by-step explanation:

Given that,

p(x) = $x^3 - 3x^2 - x + 3$

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,

$p(3) = (3)^3 - 3(3)^2 - (3) + 3$

= 27 - 3 * 9

= 27 - 27

= 0

∵ (x - 3) is a factor of p(x)

by using trial and error

(x - 1) = 0

∵ x = 1

$p(1) = (1)^3 - 1(1)^3 - 1 + 1$

= 1 - 1

= 0

∵ (x - 1) is a factor of p(x)

Thus,

p(x) =  $x^3 - 3x^2 - x + 3$

= $x^2 (x - 3) - 1 (x - 3)$

= $(x - 3) (x^2 - 1)$

= $(x - 3) (x^2 - 1^2)$

$= (x - 3) [(x + 1) (x - 1)]$

= $= (x - 3) (x + 1) (x - 1)$

Learn more: Factorize

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