Question

find a cubic polynomial whose zeros are -2,-3,-1

Answer 1

x^3-(-2-3-1)x^2+(-2*-3-3*-1-1*-2)x - (-2*-3*-1)

=x^3+6x^2 +(6+3+2)x + 6

=x^3 + 6x^2 + 11x + 6

ans.

Answer 2

Answer:

∝ = -3, β = -2, r = -1

= ∝ + β + r = (3) + (-2) + (1)

= (-3) (-2) + (-1)

= (4) - (2)

= 2

= ∝β + βr + ∝r = -(3) (-2) + (-1) + (-3) (-1)

= -6 + (-2) + -3

= -8 + (-3)

= 5

= ∝βr = (-3) (-2) (-1)

= -5

= k [x³ - (∝ + β + r)  + (∝β + βr + ∝r)  - ∝β]

=  k [x³ - (2)  + (5) - (-5)

=  k [x³ - 2) 5 + 5

= k = 1, p (x) = x² + 2 - 5 + 6

= x  = 1 => f(1) = (1)³ - 2(1)² - 5(1) + 6

= 1 - 2 - 5 + 6

= +4 -4

= 0

∴ (x -2) is a factor of p (x) = x³ + 2 - 5x + 6

  x - 2 ) x³ + 2x² - 5x + 6 ( x² + 4 + 3

            x³ + 2x²

-----------------------------------------------------

                +4x² - 5x

                 -4x² + 8x

-----------------------------------------------------

                             3x + 6

                             3x + 6

-----------------------------------------------------

                                  0

-----------------------------------------------------

= (x²+ 4x + 3)

= x² - x - 3x + 3

= x (x - 1) -3 (x - 1)

= (x - 3) (x -3) = 0

= (x - 3) = 0

= x = 3

= (x -1) = 0

= x = 1