Please help me?

Question

The cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x) = 0 are 1, k and k^2. It is given that f(x) has a remainder of 7 when divided by x - 2.

(i) Show that k^3 - 2k^2 - 2k - 3 = 0

(ii) Hence find a value for k and show that there are no other real values of k which satisfy this equation.

Please help me...I really really have no idea on how to do...

Answer 1

(i)
From the roots, cubic polynomial can be obtained as
(x - 1)(x - k)(x - k^2)
If remainder is 7 on dividing by 2,
(2 -1)(2 - k)(2 - k^2) = 7
=> k^3 - 2k^2 - 2k - 3 = 0.

(ii)
k = 3 satisfies the equation. ( Trying 1, 2, 3, -1, -2, -3)
=> k^3 - 3k^2 + k^2 - 3k + k - 3 = 0
=> k^2(k - 3) + k(k -3) + (k -3) = 0
=> (k -3)(k^2 + k + 1) = 0
=> k-3 = 0 or k^2 + k + 1 = 0
=> k = 3 (No other real solution is possible as the other equation has imaginary roots.)

Answer 2

First, note that since f(x) is cubic, and has as roots 1, k, and k², we have that:

f(x) = c (x-1) (x-k) (x-k²)

But since the coefficient of x³ is 1, we have that c=1. So we have:

f(x) = (x-1) (x-k) (x-k²)

Since we have that f(x) has a remainder of 7 when divided by x-2, it follows from the remainder theorem that f(2) = 7. So we have:

7 = f(2) = (2-1) (2-k) (2-k²)
7 = 4 - 2k - 2k² + k³
k³ - 2k² - 2k - 3 = 0

So part (i) is done. For part 2, we simply solve this cubic for k. We know that from the rational root theorem, if k is rational then k=±1 or ±3. Quickly testing 1 and -1 yields no results, let us try 3:

3| 1 -2 -2 -3
....... 3 ..3 .3
------------------
....1 1 ...1 0

Success! k=3 is a root, and the quotient is k²+k+1. Suppose there are any real solutions other than 3. Then they are roots of k²+k+1, but the discriminant of k²+k+1 is 1-4 = -3 < 0, so that polynomial has no real roots. Thus the sole possible real value of k is k=3. And we are done.

As a side note, here is f(x):

f(x) = (x-1) (x-k) (x-k²)
f(x) = (x-1) (x-3) (x-9)
f(x) = (x² - 4x + 3) (x-9)
f(x) = x³ - 13x² + 39x - 27

Answer 3

Since the roots of f(x) = 0 are 1, k and k^2.
f(x) = (x-1)(x-k)(x-k^2)

Since f(x) has a remainder of 7 when divided by x - 2.
f(2) = 7 = (2-1)(2-k)(2-k^2)

==>k^3 - 2k^2 - 2k - 3 = 0 (ans)

(Since they say hence, we need to use the result of the previos part , which is k^3 - 2k^2 - 2k - 3 = 0)

so let g(k)=k^3 - 2k^2 - 2k - 3
when k=3
g(3) = 27-18-6-3 =0, so k-3 is a factor.
by long division, (or synthetic division if you've learn it.)
(k^3 - 2k^2 - 2k - 3 )/(k-3) = k^2+k+1
so g(k) = (k-3)(k^2+k+1) = 0
so k = 3 (ans) or k^2+k+1 = 0
for k^2+k+1 = 0, b^2-4ac = 1^2 -4(1)(1) = -3 < 0, there is no real root for this. (ans)

Answer 4

f(x) = x^3 + ax^2 +bx +c
f(1) = 1+a +b+c =0
f(k) = k^3 +ak^2 +bk +c =0
f(k^2) = k^6+ak^4 +bk^2 +c =0

Also f(2)-7=0
=> 8+4a +2b+c-7=0
=> 4a + 2b +c +1 =0
I think now ou can manipulate things to get answer

Answer 5

of course you don't know anything


you're a woman