Determine the value(s), if any, of k for which the function f(x) = x^3 + 6x^2 + kx -4 gives the same remainder when divided by x − 1 and x + 2.
first divide I get
x^2 + 7 x +(7+k) with R = 3+k
second divide I get
x^2 + 4 x + (k-8) with R = -2k+12
so
k+3 = -2k+12
3 k = 9
k = 3
To determine the value(s) of k for which the function gives the same remainder when divided by x − 1 and x + 2, we can use the remainder theorem.
The remainder theorem states that if you divide a polynomial f(x) by x - c, the remainder is equal to f(c).
In this case, we need to find the value(s) of k for which the remainder is the same when divided by both x - 1 and x + 2.
Let's start by finding the remainder of f(x) when divided by x - 1.
To do this, substitute x = 1 into f(x) and find the value:
f(1) = 1^3 + 6(1)^2 + k(1) - 4
= 1 + 6 + k - 4
= k + 3
Now, let's find the remainder of f(x) when divided by x + 2.
To do this, substitute x = -2 into f(x) and find the value:
f(-2) = (-2)^3 + 6(-2)^2 + k(-2) - 4
= -8 + 24 - 2k - 4
= 12 - 2k
Now, since we want the remainder to be the same for both x - 1 and x + 2, we can set the two expressions equal to each other:
k + 3 = 12 - 2k
Simplifying, we can combine like terms:
3k = 9
Dividing both sides by 3:
k = 3
Therefore, the value of k for which the function gives the same remainder when divided by x - 1 and x + 2 is k = 3.