# Thread: Polynomials questions again... part 2

1. ## Polynomials questions again... part 2

3. Given that $\displaystyle f(x)=x^n-nx+n-1$ for integer n>1. By considering f(x) and f '(x), show that $\displaystyle f(x)=(x-1)^2 g(x)$ is true for all polynomials g(x) with integer coefficients. Hence or otherwise
a) Show that $\displaystyle 3^{2n}-8n-1$ is divisible by 64 for all integers n>1
b) Show that the equation $\displaystyle x^4-4x+3=0$ does not have any real root other than 1

4. If $\displaystyle 3+\sqrt{3}$ is a root of the equation $\displaystyle 2x^2+3bx+3c=0$ where b and c are constants. Find the value of bc.

4. bc=-16

2. I can't help you with 1 and 2, I hate those kind of questions.

For 3, take the derivative of that function, find the critical points, and then you can show that 1 is the only time the function touch the x axis.

For 4, you know one root is $\displaystyle 3+ \sqrt 3$, the other solution is $\displaystyle 3- \sqrt 3$, substitute these values to x, you get 2 equations, solve the simultaneous equations, and you'll get your b and c value.

Another way of doing it is with the root coefficient relationship. In the quadratic equation $\displaystyle ax^2+bx+c$, suppose the roots are $\displaystyle \alpha$ and $\displaystyle \beta$, $\displaystyle \alpha + \beta = -b, \alpha \beta = c$

3. Originally Posted by cloud5
4. If $\displaystyle 3+\sqrt{3}$ is a root of the equation $\displaystyle 2x^2+3bx+3c=0$ where b and c are constants. Find the value of bc.
If $\displaystyle 3+\sqrt{3}$ is a root, then so is $\displaystyle 3-\sqrt{3}$.
For a quadratic in the form $\displaystyle f(x) = ax^2 + bx + c$, the sum of its roots would be $\displaystyle -\frac{b}{a}$, and the product of its roots would be $\displaystyle \frac{c}{a}$. Plug in (be careful that the b and c are not the same as the b and c in your problem, ie. a = 2, "b" = 3b, and "c" = 3c):
\displaystyle \begin{aligned} (3 + \sqrt{3}) + (3 - \sqrt{3}) &= -\frac{3b}{2} \\ 6 &= -\frac{3b}{2} \\ 12 &= -3b \\ b &= -4 \end{aligned}

\displaystyle \begin{aligned} (3 + \sqrt{3})(3 - \sqrt{3}) &= \frac{3c}{2} \\ 9 - 3 &= \frac{3c}{2} \\ 6 &= \frac{3c}{2} \\ 12 &= 3c \\ c &= 4 \end{aligned}

So bc = (-4)(4) = -16.

01

4. Originally Posted by chengbin
Another way of doing it is with the root coefficient relationship. In the quadratic equation $\displaystyle ax^2+bx+c$, suppose the roots are $\displaystyle \alpha$ and $\displaystyle \beta$, $\displaystyle \alpha + \beta = -b, \alpha \beta = c$
This is only true if the leading coefficient is one, or $\displaystyle {\color{red}x^2} + bx + c = 0$.

01

5. Hello, cloud5!

4. $\displaystyle 3+\sqrt{3}$ is a root of the equation $\displaystyle 2x^2+3bx+3c\:=\:0$ where $\displaystyle b$ and $\displaystyle c$ are constants.
Find the value of bc.

Answer: $\displaystyle bc=-16$
We expected to be familiar with this fact . . .

Given: .$\displaystyle x^2 + px + q \:=\:0$, a quadratic equation with leading coefficient 1.

. . If the roots are $\displaystyle \alpha$ and $\displaystyle \beta$, then: .$\displaystyle \begin{Bmatrix}\alpha + \beta &=& \text{-}p \\ \alpha\beta &=& q \end{Bmatrix}$

In English: .$\displaystyle \begin{array}{ccc}\text{Sum of the roots} &=& \text{negative of the }x\text{-coefficient.} \\ \text{Product of the roots} &=& \text{the constant term.} \end{array}$

We have: .$\displaystyle x^2 + \frac{3b}{2}x + \frac{3c}{2} \;=\;0$

If one root is $\displaystyle (3 + \sqrt{3})$, the other root is $\displaystyle (3-\sqrt{3})$
. . The roots appear in "conjugate pairs."

The sum of the roots is $\displaystyle -\frac{3b}{2}\!:\quad (3+\sqrt{3}) + (3 - \sqrt{3}) \:=\:-\frac{3b}{2}$
. . Hence: .$\displaystyle -\frac{3b}{2} \:=\:6 \quad\Rightarrow\quad\boxed{ b \:=\:-4}$

The product of the roots is $\displaystyle \frac{3c}{2}\!:\quad (3+\sqrt{3})(3-\sqrt{3}) \:=\:\frac{3c}{2}$
. . Hence: .$\displaystyle \frac{3c}{2} \:=\:6 \quad\Rightarrow\quad\boxed{ c \:=\:4}$

Therefore: .$\displaystyle bc \;=\;(-4)(4) \;=\;{\color{blue}-16}$

6. Originally Posted by cloud5
3. Given that $\displaystyle f(x)=x^n-nx+n-1$ for integer n>1. By considering f(x) and f '(x), show that $\displaystyle f(x)=(x-1)^2 g(x)$ is true for all polynomials g(x) with integer coefficients.
First, the way this is written, it is NOT true! That should be "for some polynomial g(x)". Obviously $\displaystyle (x-1)^2g(x)$ cannnot always be equal to f(x) no matter what g is! What you are really asked to prove is that f(x) has 1 as a double zero. Don't look at f(x) and f'(x), look at f(1) and f'(1).

Hence or otherwise
a) Show that $\displaystyle 3^{2n}-8n-1$ is divisible by 64 for all integers n>1
If $\displaystyle x= 9= 3^2$, $\displaystyle f(9)= 9^n- 9n+ n- 1= 3^{2n}- n- 1$ so $\displaystyle f(x)= (x-1)g(x)$. Now, what does that look like when x= 3?

b) Show that the equation $\displaystyle x^4-4x+3=0$ does not have any real root other than 1
This is $\displaystyle f(x)= x^4- 4x+ 4- 1$ and so can be written as [tex](x-1)^2g(x)[tex] for some polynomial g. You can use polynomial division by $\displaystyle (x-1)^2= x^2- 2x+ 1$ (or synthetic division twice) to find g, a quadratic equation. What are the roots of g?

4. If $\displaystyle 3+\sqrt{3}$ is a root of the equation $\displaystyle 2x^2+3bx+3c=0$ where b and c are constants. Find the value of bc.