Question
Using factor theorem, factorize the following polynomials: $x^3 - 6x^2 + 3x + 10$
✓
Answer
Let $x = 2 f(2) = 2^3 + 6(2)^2 + 3(2) + 10 = 8 - 24 + 6 + 10 = 0$
$\therefore x = 2$ is a solution $f(x)$ i. e $(x - 2)$ is a factor of $f(x)$
By division algorithm $x^3 - 6x^2 + 3x + 10$
$= (x - 2)(x^2 - 4x - 5)$
$= (x - 2)(x^2 - 5x + x - 5)$
$= (x - 2)(x(x - 5) + 1(x - 5))$
$= (x - 2)(x - 5)(x + 1)$
$\therefore x^3 - 6x^2 + 3x + 10 = (x - 2)(x - 5)(x + 1)$
$\therefore x = 2$ is a solution $f(x)$ i. e $(x - 2)$ is a factor of $f(x)$
By division algorithm $x^3 - 6x^2 + 3x + 10$
$= (x - 2)(x^2 - 4x - 5)$
$= (x - 2)(x^2 - 5x + x - 5)$
$= (x - 2)(x(x - 5) + 1(x - 5))$
$= (x - 2)(x - 5)(x + 1)$
$\therefore x^3 - 6x^2 + 3x + 10 = (x - 2)(x - 5)(x + 1)$
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