if (x – 1), (x + 1) and (x – 2) are factors of the polynomial ax\(^3\) + bx\(^2\) + cx – 1, find a, b, c in that order.
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-1/2, 1., 1/2
- 1/2, 1, 1/2
- 1/2, 1, -1/2
- 1/2, -1, 1/2
The correct answer is: A
Explanation
This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2.
Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2
Equate each factor to zero to get the values of x
x-1 = 0, x = 1
x +1 = 0, x = -1
x - 2 = 0, x = 2
substitute these value of x in the given equation
\(ax^3 + bx^2\) + cx - 1
when x = 1
\(ax^3 + bx^2\) + cx - 1 = a + b + c - 1= 0 ------------------------(1)
when x = -1
\(ax^3 + bx^2\) + cx - 1 = -a + b - c - 1 = 0-------------------------(2)
when x = 2
\(ax^3 + bx^2\) + cx - 1 = 8a + 4b + 2c - 1 = 0 ----------------------(3)
add equations 1 and 2
a + b + c - 1= 0
+ -a + b - c - 1 = 0
b = 1
put b = 1 into equation (1)
a + b + c - 1= 0 = a + 1 + c -1 = 0
a = - c
put a = - c into equation(3)
8a + 4b + 2c - 1 = 0,
-8c + 4 + 2c - 1 = 0 ( b = 1 remember)
-6c = -3
c = \(\frac{-3}{-6}\) = \(\frac{1}{2}\)
put c = \(\frac{1}{2}\) and b = 1 into equation (1)
a + b + c - 1= 0
a + 1 + \(\frac{1}{2}\) - 1 = 0
a + \(\frac{1}{2}\) = 0
then a = \(\frac{-1}{2}\)
There is an explanation video available .