Find a cubic polynomial whose zeroes are -1, 2 and 3
Answers
Step-by-step explanation:
Since, -1,2 & 3 are the Zeros
(x+1),(x-2)&(x-3) must be the factors.
Therefore, Multiplying we get
f(x) = (x+1)(x²-5x+6)
f(x)= x³-5x²+x+6
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 - 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