Find a cubic polynomial with zeros 3, 2 and-1
Answers
Hello friend
Let the cubic polynomial be p(x)
It's zeros are 3,2,-1
So the factors will be (x-3),(x-2),(x+1)
A cubic polynomial will at most have three zeros
So by multiplying all its three factors, we will get the polynomial
(x-3)(x-2)(x+1)
=(x²-2x-3x+6)(x+1)
=(x²-5x+6)(x+1)
=x³+x²-5x²-5x+6x+6
=x³-4x²+x+6
Here we got the polynomial
Hope it helps!
Brainiest answer please.
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