A function, f , is given byf (x) = x3 + 2x2 + x + k,where k is a constant and, when f (x) is divided by x − 1, the remainder is 4.Find k and hence find the solutions of the equation f (x) = 0.
Question
A function, f , is given byf (x) = x3 + 2x2 + x + k,where k is a constant and, when f (x) is divided by x − 1, the remainder is 4.Find k and hence find the solutions of the equation f (x) = 0.
Solution
To find the value of k, we can use the Remainder Theorem which states that if a polynomial f(x) is divided by (x-a), the remainder is f(a). In this case, x - 1 is the divisor and the remainder is 4. So, we substitute x = 1 into the function:
f(1) = (1)^3 + 2*(1)^2 + 1 + k = 4 => 1 + 2 + 1 + k = 4 => k = 4 - 4 = 0
So, the constant k is 0.
Now, we substitute k = 0 into the function f(x) to find the solutions of the equation f(x) = 0:
f(x) = x^3 + 2x^2 + x = 0
This is a cubic equation and can be factored as follows:
x(x^2 + 2x + 1) = 0 => x(x + 1)^2 = 0
Setting each factor equal to zero gives the solutions of the equation:
x = 0 or x = -1
So, the solutions of the equation f(x) = 0 are x = 0 and x = -1.
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