Question

find cubic polynomial whos zero are 1,-2 and 3

Answer 1

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)

= -6

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

=  k [x³ - (2)  + (-5) - (-6)

=  k [x³ - 2) - 5 + 6

= k = 1, p (x) = ² + 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

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                             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