Find the Zeros of the Cubic Polynomial using Trial and Error method.
(i) f(x) = x³ - 3x² + 5x - 3
(ii) f(x) = 2x³ + x² + 5x - 2
Answers
I think you should first apply hidden trial and than divide it after which you have to solve the bracket. After which you will get answer.
Answer:
(i) f(x) = x3 = 3x2 +5x – 3
= 1 – 3 + 5 – 3
= 6 – 6
= 0
∴ (x-1) is a factor of f(x) = x3 – 3x2 + 5x – 3.
x – 1 ) x3 – 3x2 + 5x – 3 ( x2 – 2x + 3
x3 – x2
------------------------------------------------------
-2x2 + 5x
-2x2 + 2x
------------------------------------------------------
+3x - 3
-3x + 3
-------------------------------------------------------
0
= x2 – 2x + 3
= x = -b + √d/2a
= - (-2) + √6/2(1)
= 2 + 4/2, 2 - 4/2
= 6/2, -2/2
= 3, -1
(ii) f(x) = 2x³ + x² + 5x - 2
= x = -2 => f (-2)³ + (-2)² - 5 (-2) + 2
= 2 (-8) + 4 + 10 + 2
= -16 + 14 + 2
= -16 + 16
= 0
∴ (x + 2) is a factor of f(x) = 2x³ + x² + 5x - 2
x + 2 ) 2x³ + x² + 5x - 2 ( 2x² - 3x + 1
2x³ + 4x²
----------------------------------------------------
-3x² - 5x
-3x² - 6x
----------------------------------------------------
x - 2
x - 2
----------------------------------------------------
0
----------------------------------------------------
= 2x² - 3x + 1
= 2x² - 2x - x + 1
= 2x (x - 1) -1 (x - 1)
= (2x - 1) (x - 1) = 0
= (x - 1) = 0
= x = 1
= (2x - 1) = 0
= 2x = 1
= x = 1/2
∴ 1/2, 1 , -2 are the Zeros.