solve x^3 - 4x^2 - 3x + 18 <= 0

Answer 1

Write:

- 4 x² as 2 x² - 6 x²

and

- 3 x as - 12 x + 9 x

x³ - 4 x² - 3 x + 18 = x³ + 2 x² - 6 x² - 12 x + 9 x + 18 =

( x³ + 2 x² ) ( - 6 x² - 12 x ) + ( 9 x + 18 ) =

x² ∙ ( x + 2 ) - 6 x ∙ ( x + 2 ) + 9 ∙ ( x + 2 ) = ( x + 2 ) ∙ ( x² - 6 x + 9 )

Since ( a - b )² = a² - 2 ∙ a ∙ b + b²

( x - 3 )² = x² - 2 ∙ x ∙ 3 + 3² = x² - 6 x + 9

So

x³ - 4 x² - 3 x + 18 = ( x + 2 ) ∙ ( x² - 6 x + 9 ) = ( x + 2 ) ∙ ( x - 3 )²

( x + 2 ) ∙ ( x - 3 )² ≤ 0

( x - 3 )² is always positive so

( x + 2 ) ∙ ( x - 3 )² is less of zero when

x + 2 < 0

x < - 2

( x + 2 ) ∙ ( x - 3 )² is equal zero 0 when

x + 2 = 0 and x - 3 = 0

x = - 2 and x = 3

x³ - 4 x² - 3 x + 18 ≤ 0

The solutions are:

x < - 2 , x = 2 , x = 3

x ≤ - 2 , x = 3