Question
Using factor theorem, factorize the following polynomials: $x^3 + 2x^2 - x - 2$
✓
Answer
Let $x = 1 f(1) = 1^3 + 2(1)^2 - 1 - 2 = 0$
$\therefore x = 1$ is a solution
$\Rightarrow x - 1 = 0$ i. $e (x - 1)$ is a factor of $f(x)$
By division algorithm $x^3 + 2x^2 - x - 2$
$= (x - 1)(x^2 + 3x + 2)$
$= (x - 1)(x^2 + 2x + x + 2)$
$= (x - 1)(x(x + 2) + 1(x + 2))$
$= (x - 1)(x + 2)(x + 1)$
$\therefore x^3 + 2x^2 - x - 2$
$= (x - 1)(x + 2)(x + 1)$
$\therefore x = 1$ is a solution
$\Rightarrow x - 1 = 0$ i. $e (x - 1)$ is a factor of $f(x)$
By division algorithm $x^3 + 2x^2 - x - 2$
$= (x - 1)(x^2 + 3x + 2)$
$= (x - 1)(x^2 + 2x + x + 2)$
$= (x - 1)(x(x + 2) + 1(x + 2))$
$= (x - 1)(x + 2)(x + 1)$
$\therefore x^3 + 2x^2 - x - 2$
$= (x - 1)(x + 2)(x + 1)$
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