Question

${3x}^{3} - {10x}^{2} + 9x - 2 = 0$

Answer 1

3 x^3 - 1 0 x^2 + 9 x - 2 = 0

3 x^3 - 3 x^2 - 7 x^2 + 9 x - 2 = 0

3 x^2 ( x - 1 ) - 7 x^2 + ( 7 + 2 ) x - 2 = 0

3 x^2 ( x - 1 ) - 7 x^2 + 7 x + 2 x - 2 = 0

3 x^2 ( x - 1 ) - 7 x ( x - 1 ) + 2 ( x - 1 ) = 0

( x - 1 ) ( 3 x^2 - 7 x + 2 ) = 0

( x - 1 ) { 3 x^2 - ( 6 + 1 ) x + 2 } = 0

( x - 1 ) { 3 x^2 - 6 x - x + 2 } = 0

( x - 1 ) { 3 x( x - 2 ) - ( x - 2 ) } = 0

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

Hence,

x = 1

Or, x = 2

Or, x = 1 / 3

Answer 2

уσυя αиѕωєя !!

υѕιиg fα¢тσя тнєσяєм !

$= >3 {x}^{3} - 10 {x}^{2} +9x - 2 = 0 \\ \\ = > 3 {x}^{3} + 3 {x}^{2} - 7 {x}^{2} + 7x + 2x - 2 = 0 \\ \\ = > 3 {x}^{2} (x - 1) - 7x(x - 1) + 2(x - 1) = 0 \\ \\ = > (x - 1)(3 {x}^{2} - 7x + 2) = 0 \\ \\ = > (x - 1)(3 {x}^{2} - 6x - 1x + 2) = 0 \\ \\ = > (x - 1)(3x(x - 2) - 1(x - 2)) = 0 \\ \\ = > (x - 1)(3x - 1)(x - 2) = 0$

нєи¢є, ωє gσт тняєє ναℓυєѕ fσя χ

тнαт αяє:

=> χ - 1 = 0

=> χ = 1
_________

=> 3χ - 1 = 0

=> 3χ = 1

=> χ = 1/3
_________

=> χ - 2 = 0

=> χ = 2

тнαикѕ !!