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