Question

find zeroes of 6x²-3-7x​

Answer 1

Step-by-step explanation:

6x^2-7x-3=0

6x^2-9x+2x-3=0

3x(2x-3)+1(2x-3)=0

(3x+1)(2x-3)=0

hence the zeros of the polynomial are. x = -1/3, 3/2

Answer 2

Answer:

$6 {x}^{2} - 3 - 7x$

$= 6 {x}^{2} - 7x - 3$

$= 6 {x}^{2} + 2 x - 9x - 3$

$= 2x(3x + 1) - 3(3x + 1)$

$= (3x + 1)(2x - 3)$

ZEROES :–

$3x + 1 = 0$

$3x = -1$

$x = \frac{-1}{3}$

$2x - 3 = 0$

$2x = 3$

$x = \frac{3}{2}$

$x = \frac{-1}{3} \: and \: \frac{3}{2}$