# How do you use the factor theorem to determine whether x-2 is a factor of 4x^3 – 3x^2 – 8x + 4?

Jun 8, 2018

see below

#### Explanation:

the factor theorem states

$\left(x - a\right) \text{ is a factor of } f \left(x\right) \iff f \left(a\right) = 0$

$f \left(x\right) = 4 {x}^{3} - 3 {x}^{2} - 8 x + 4$

$\text{we have } \left(x - 2\right) \implies a = 2$

$f \left(2\right) = 4 \times {2}^{3} - 3 \times {2}^{2} - 8 \times 2 + 4$

$f \left(2\right) = 4 \times 8 - 3 \times 4 - 16 + 4$

$f \left(2\right) = 32 - 12 - 16 + 4$

$f \left(2\right) = 8 \ne 0$

$\therefore x - 2 \text{ is not a factor of } 4 {x}^{3} - 3 {x}^{2} - 8 x + 4$

however by the remainder theorem when

$f \left(x\right) = 4 {x}^{3} - 3 {x}^{2} - 8 x + 4 \text{ is divided by "(x-2) " the remainder is } 8$

the factor theorem being a special case of the remainder theorem

Jun 8, 2018

$\text{not a factor}$

#### Explanation:

$\text{if "x-2" is a factor then } f \left(2\right) = 0$

$4 {\left(2\right)}^{3} - 3 {\left(2\right)}^{2} - 8 \left(2\right) + 4 = 8$

$\text{hence "x-2" is not a factor of } 4 {x}^{3} - 3 {x}^{2} - 8 x + 4$