Given that ( )f x x kx p4 3= + + is exactly divisible by x 2+ and ( )f xl is exactly divisible byx2 1- , find the value of k and of p. [4](ii) Using the values of k and p found in part (i), show that ( ) ( ) ( )f x x ax bx c2 2= + + + ,where a, b and c are integers to be found. [2](iii) Hence show that ( )f x 0= has only one solution and state this solution. [2]
Question
Given that ( )f x x kx p4 3= + + is exactly divisible by x 2+ and ( )f xl is exactly divisible byx2 1- , find the value of k and of p. 4 Using the values of k and p found in part (i), show that ( ) ( ) ( )f x x ax bx c2 2= + + + ,where a, b and c are integers to be found. 2 Hence show that ( )f x 0= has only one solution and state this solution. [2]
Solution
(i) Given that f(x) = x^4 + kx^3 + px is exactly divisible by x^2 + 1 and f(x) is exactly divisible by x^2 - 1, we can use the factor theorem to find the values of k and p.
The factor theorem states that if a polynomial f(x) is divisible by x - a, then f(a) = 0.
So, if f(x) is divisible by x^2 + 1, then f(i) = 0 and f(-i) = 0, where i is the imaginary unit.
Substituting x = i and x = -i into the equation gives:
f(i) = i^4 + ki^3 + pi = 0 f(-i) = (-i)^4 + k(-i)^3 + p(-i) = 0
Solving these two equations simultaneously gives k = 0 and p = 0.
(ii) Substituting k = 0 and p = 0 into the equation f(x) = x^4 + kx^3 + px gives f(x) = x^4, which can be rewritten as f(x) = x^2 * x^2 = (x^2 + 0x + 0)^2. So, a = 1, b = 0, and c = 0.
(iii) The equation f(x) = 0 becomes x^4 = 0. The only solution to this equation is x = 0.
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