Question

a) x⁴-x³-13x²+x+12=0;
б) x⁴-x³-7x²+x+6=0.​

Answer 1

Ответ:

a) x⁴ - x³ - 13x² + x + 12 = 0

x⁴ - x³ - 13x² + 13x - 12x + 12 = 0

x³ * ( x - 1 ) - 13x * ( x - 1 ) - 12 * (x - 1) = 0

( x - 1 ) * ( x³ - 13x - 12 ) = 0

( x - 1 ) * ( x³ - x - 13x - 12 + x² - x² ) = 0

( x - 1 ) * ( x² * ( x + 1 ) - x * ( x + 1 ) - 12 * ( x + 1) ) = 0

( x - 1 ) * ( x + 1 ) * ( x² - x - 12) = 0

( x - 1 ) * ( x + 1 ) * ( x² + 3x - 4x - 12) = 0

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

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

x - 1 = 0

x + 1 = 0

x + 3 = 0

x - 4 = 0

x = 1

x = - 1

x = - 3

x = 4

Ответ = 1, -3, -1, 4.