1. ## more constructing polynomials.

If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

AND

Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.

2. Originally Posted by THSKluv
If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

AND

Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.
1. For the polynomial $\displaystyle P(x) = 2x^4 + 8x^3 + px^2 + qx + r$, if $\displaystyle x = -1, x = 1, x = 3$ are roots of the polynomial, then $\displaystyle (x +1), (x - 1), (x - 3)$ are factors of the polynomial.

So that means, by the factor and remainder theorems...

$\displaystyle P(-1) = 0, P(1) = 0, P(3) = 0$.

So...
$\displaystyle 2(-1)^4 + 8(-1)^3 + p(-1)^2 + q(-1) + r = 0$

$\displaystyle 2 - 8 + p - q + r = 0$

$\displaystyle -6 + p - q + r = 0$

$\displaystyle p - q + r = 6$.

$\displaystyle 2(1)^4 + 8(1)^3 + p(1)^2 + q(1) + r = 0$

$\displaystyle 2 + 8 + p + q + r = 0$

$\displaystyle 10 + p + q + r = 0$

$\displaystyle p + q + r = -10$.

$\displaystyle 2(3)^4 + 8(3)^3 + p(3)^2 + q(3) + r = 0$

$\displaystyle 162 + 216 + 9p + 3q + r = 0$

$\displaystyle 378 + 9p + 3q + r = 0$

$\displaystyle 9p + 3q + r = -378$.

Now you have three equations in three unknowns which you can solve simultaneously for $\displaystyle p, q, r$.

$\displaystyle p - q + r = 6$

$\displaystyle p + q + r = -10$

$\displaystyle 9p + 3q + r = -378$.

3. Originally Posted by THSKluv
If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

AND

Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.
2. To show that $\displaystyle x^4 + x^3 + x + 1 = 0$ has a double root at $\displaystyle x = -1$, you have to show that $\displaystyle (x + 1)^2$ is a factor.

$\displaystyle x^4 + x^3 + x + 1 = 0$

$\displaystyle x^3(x + 1) + 1(x + 1) = 0$

$\displaystyle (x + 1)(x^3 + 1) = 0$

$\displaystyle (x + 1)(x + 1)(x^2 - x + 1) = 0$

$\displaystyle (x + 1)^2(x^2 - x + 1) = 0$.

4. 1) If -1, 1, 3 are the roots then

$\displaystyle \left\{\begin{array}{lll}P(-1)=0\\P(1)=0\\P(3)=0\end{array}\right.\Rightarrow\ left\{\begin{array}{lll}p-q+r=6\\p+q+r=-10\\9p+3q+r=-378\end{array}\right.$

Substracting the first equation from the second we have $\displaystyle 2q=-16\Rightarrow q=-8$

Replacing q in the first and the third equation we have

$\displaystyle \left\{\begin{array}{ll}p+r=-2\\9p+r=-354\end{array}\right.$

Substracting the first from the second we get $\displaystyle 8p=-352\Rightarrow p=-44\Rightarrow r=42$

2) $\displaystyle x^4+x^3+x+1=c^3(x+1)+x+1)=(x+1)(x^3+1)=(x+1)^2(x^2-x+1)$

Then $\displaystyle x_1=x_2=-1$ and $\displaystyle x^2-x+1=0$ has no real roots.

5. Originally Posted by red_dog
1) If -1, 1, 3 are the roots then

$\displaystyle \left\{\begin{array}{lll}P(-1)=0\\P(1)=0\\P(3)=0\end{array}\right.\Rightarrow\ left\{\begin{array}{lll}p-q+r=6\\p+q+r=-10\\9p+3q+r=-378\end{array}\right.$

Substracting the first equation from the second we have $\displaystyle 2q=-16\Rightarrow q=-8$

Replacing q in the first and the third equation we have

$\displaystyle \left\{\begin{array}{ll}p+r=-2\\9p+r=-354\end{array}\right.$

Substracting the first from the second we get $\displaystyle 8p=-352\Rightarrow p=-44\Rightarrow r=42$

2) $\displaystyle x^4+x^3+x+1=c^3(x+1)+x+1)=(x+1)(x^3+1)=(x+1)^2(x^2-x+1)$

Then $\displaystyle x_1=x_2=-1$ and $\displaystyle x^2-x+1=0$ has no real roots.
How do you get $\displaystyle 9p + 3q + r = -378$?

6. Originally Posted by Prove It
How do you get $\displaystyle 9p + 3q + r = -378$?
Replace $\displaystyle x=3$:

$\displaystyle 2\cdot3^4+8\cdot 3^3+p\cdot3^2+q\cdot 3+r=0$

$\displaystyle 162+216+9p+3q+r=0\Rightarrow 9p+3q+r=-378$

7. Originally Posted by red_dog
Replace $\displaystyle x=3$:

$\displaystyle 2\cdot3^4+8\cdot 3^3+p\cdot3^2+q\cdot 3+r=0$

$\displaystyle 162+216+9p+3q+r=0\Rightarrow 9p+3q+r=-378$
Thanks. I had a small error - my arithmetic is well-known to be shocking.

8. Originally Posted by THSKluv
If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""
The sum of the roots is -4 not 4, so the missing root is -7.

CB