Question

Show that -1 and 2/3 are the zeroes of polynomial 3x^3 - 5x^2 - 4x + 4. Also, find the third zero.

PLEASE ANSWER FAST !!

Answer 1

Answer:

given polynomial

$3 {x}^{3} - 5 {x}^{2} - 4x + 4$

let x= -1

$3{(-1)}^{3}-5{(-1)}^{2}-4(-1)+4$

$=> - 3 - 5 + 4 + 4$

$=> - 8 + 4 = 0$therefore -1 is the zero of the polynomial

let x=2/3

$3 (\frac{2}{3})^{3} - 5 ({ \frac{2}{3} })^{2} - 4( \frac{2}{3}) + 4$

$=> 3 \times \frac{8}{27} - 5 \times \frac{4}{9} - 4 \frac{2}{3} + 4$

$= > \frac{8}{9} - \frac{20}{9} - \frac{8}{3} + 4$

$= > \frac{8 - 20 - 24 + 36}{9}$

$= > \frac{44 - 44}{9} = 0$

therefore x=2/3 is also the zeros of the polynomial

to find third zero :-

since -1 and 2/3 are the zeros of the polynomial

therefore(x+1) and (3x-2) are two factors of the polynomial

$3{x}^{3}-5{x}^{2}-4x+4\div(x+1)(3x -2)$

$= > \frac{3 {x}^{3} - 5 {x}^{2} - 4x + 4}{(3x - 2 )(x + 1)}$

$=>\frac{(x + 1)(3 {x}^{2} - 8x + 4)}{(x + 1)(3x -2) }$

$= > \frac{(3 {x}^{2} - 8x + 4) }{(3x - 2 )}$

$= > \frac{3 {x}^{2} - 6 x - 2x + 4 }{3x -2 }$

$= > \frac{3x(x - 2) - 2(x - 2) }{3x -2}$

$= > \frac{(x - 2)(3x - 2)}{3x - 2}$

$= > (x - 2)$

therefore( x-2) is a factor of polynomial

x - 2 = 0

x = 2 is a third zeros of the polynomial