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