Question

using factor theorem factorize
$x ^ 3 - 3x ^ 2 - x + 3$
​

Answer 1

Answer:

1 ³ - 3 * 1 ² - 1 + 3

1 - 3 - 1 + 3

= 0

x - 1

Factor

x = 1

Answer 2

Explanation:

Let the given polynomial be p(x).

Through the method of trial and error,it can be found that when x is substituted with 1, p(1) = 0.

$p(1) = {1}^{3} - 3(1) {}^{2} - 1 + 3$

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

$p(1) = 0$

1 is the zero of the polynomial x - 1. Therefore, x - 1 is a factor of p(x). Hence,

$p(x) = {x}^{2} (x - 1) - 2(x - 1) - 3(x - 1)$

$p(x) = (x - 1)( {x}^{2} - 2x - 3)$

$p(x) = (x - 1)( {x}^{2} + x - 3x - 3)$

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

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

Answer: p(x) = (x-1)(x+1)(x-3)