Question

3c³-10c²+9c-1=0
Find values of c​

Answer 1

Answer:

${3c}^{3} - {10c}^{2} + 9c - 1 = 0 \\ {3c}^{3} - {10c}^{2} + 9c - 1 = 0 \\ c( {3c}^{2} - 10c + 9- 1) = 0 \\ c( {3c}^{2} - 10c + 8) = 0 \\ c( {3c}^{2} - 12c + 2c + 8) = 0 \\ {3c}^{2} - 12c + 2c + 8 = 0 \\ - 3c(c + 4) + 2(c + 4) = 0 \\ ( - 3c + 2)(c + 4)$

either,

(-3c+2)=0

$- 3c = - 2 \\ c = \frac{ - 2}{3}$

or,

c+4=0

c=-4

I hope it's help you