Question

Using factor theorem, factorize each of the following polynomials

$x^3-6x^2$ + 3x + 10

Answer 1

Let f(x) = $x^3-6x^2$ + 3x + 10

The factors of the constant term 10 are

$\pm 1,\pm 2,\pm 5,\pm 10$,

Let x = - 1, then

f(-1) $=(-1{)}^3-6(-1{)}^2$ + 3(-1) + 10

= - 1 - 6- 3 + 10 = 10 - 10 = 0

$\therefore$ f(-1) = 0, then

x + 1 is its factor

f(-2) $=(-2{)}^3-6(-2{)}^2$ + 3(-2) +10

= - 8 - 24 - 6 + 10 = - 38 + 10 = - 28

$\therefore f(-2)\ne 0$

$\therefore$ x + 2 is not its factor

Let x = 2, then

f(2) $=(2{)}^3-6(2{)}^2+3\times$ 2 +10

= 8 - 24 + 6 + 10 = 24 - 24 = 0

$\therefore$ f(2) = 0

$\therefore$ x - 2 is a factor of f(x)

Let x = 5, then

f(5) $=(5{)}^3-6(5{)}^2+3\times$ 5 + 10

= 125 -150 + 15 + 10 = 150 - 150 = 0

$\therefore$ f(5) =0

$\therefore$ x - 5 is a factor of f(x)

f(x) = (x+1) (x-2) (x-5)