find a cubic polynomial whose zeros are -2,-3,-1
Answers
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:
∝ = -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