1. ## Equations

Hi! Thanks for helping me!

Show that the equation x^3 - x^2 +x - 21 = 0 has only one solution.

and

Without dividing find the remainder when x^3 - 4x + 12x - 2 is divided by x+1.

and one more

For what values of k does the graph of y = 2x^2 - kx + 3 have a TP on the x-axis?

2. Ok i got number 3, i realised that delta = 0 when there is a TP on x-axis.
So k = +-2 root 6.

3. Originally Posted by classicstrings
Hi! Thanks for helping me!

Show that the equation x^3 - x^2 +x - 21 = 0 has only one solution.
You can easily show it by demonstrating that this function is increasing only, and hence intersects the x-axis exactly one. Thus, it has only one real solution.

But you need to have the knowledge of calculus, I do not know if you studied it. Did you?

4. Well i know a bit of calculus as we just started but there is no other way to answer question?

5. Originally Posted by classicstrings
Well i know a bit of calculus as we just started but there is no other way to answer question?
Hello,

your equation $\displaystyle T(x)=x^3 - x^2 +x - 21 = 0$ consists of a term in x T(x) which is equal to zero. If there is a solution you can split this term into a linear factor (x-s) (s stands for solution) and the remaining term R(x). So you get $\displaystyle T(x)=(x-s) \cdot R(x)$. When you multiply the RHS of this equation you can see, that s must be a divisor of the constant summand of your equation. All integer divisors of 21 are: +-1, +-3, +-7, +-21.
Now do a little bit trial and error and you'll find pretty soon, that 3 must be a solution of your equation. That means you can divide the LHS of your equation by (x-3):
$\displaystyle \left( x^3 - x^2 +x - 21 \right)/\left(x-3 \right) = x^2+2x+7$
Because $\displaystyle x^2+2x+7 > 0$ for all $\displaystyle x \in \mathbb{R}$ there are no other possible solutions.

Greetings

EB

6. Oh i see, thanks! Yes you need to sub in factors of 21. Normally, I just sub in values starting from 1, then -1, to 2 to -2 and so on.