Question

find the zeroes of thequadratic polynomial 3x^2-x-4​

Answer 1

Answer:

$= > 3 {x}^{2} - x - 4 = 0 \\ = > 3 {x}^{2} + 3x - 4x - 4 = 0 \\ = > 3x(x + 1) - 4(x + 1) = 0 \\ = > (3x - 4)( x+ 1) = 0 \\ = > either \: 3x - 4 = 0 \: or \: x + 1 = 0 \\ = > either \: x = \frac{4}{3} \: or \: x = - 1.$

Answer 2

$\huge\bigstar\mathfrak\pink{\underline{\underline{SOLUTION:}}}$

${3x}^{2} - x - 4 \\ = > 3 {x}^{2} - x - 4 = 0 \\ \\ = > {3x}^{2} + 3x - 4x - 4 = 0 \\ \\ = > 3x(x + 1) - 4(x + 1) = 0 \\ \\ = > (x + 1)(3x - 4) = 0 \\ \\ = > x + 1 = 0 \: \: or \: 3x - 4 = 0 \\ \\ = > x = - 1 \: \: or \: \: x = \frac{4}{3}$

Hope it helps ☺️