Question

Find the zeroes of 3x2+x-4

Answer 1

3x² + x - 4 = 0

3x²+4x-3x-4= 0

3x(x-1) +4(x-1) = 0

(3x+4)(x-1) =0

x = 1 or -4/3

Hope it helps you...

Answer 2

Answer:

$p(x) = 3x {}^{2} + x - 4 \\ \: \: \: \: \: \: \: \: \: = 3x {}^{2} + (4 - 3)x - 4 \\ \: \: \: \: \: \: \: \: \: = 3x {}^{2} + 4x -3x - 4 \\ \: \: \: \: \: \: \: \: \: = x(3x - 4) - 1( 3x - 4) \\ \: \: \: \: \: \: \: \: \: = (x - 1)(3x - 4) \\ for \: zeros \: \\ p(x) = 0 \: \: \\ x - 1 = 0 \\ x = 1 \\ and \: \\ 3x - 4 = 0 \\ x = \frac{4}{3}$

Therefore , Zeros of p(x) are 1 and 4/3