+0

HEEEEELP ASAP! This is due tomorrow!!!

0
532
5

Determine a constant k such that the polynomial P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) is divisible by x+y+z

Thanks for the help!

Aug 23, 2018
edited by Guest  Aug 23, 2018

#1
+96
+2

If you assume such a constant exists, then you can solve for it by setting up an equation based on P(x, y, z) = 0 whenever x + y + z = 0. For example, P(2, -1, -1) = 0.

Aug 23, 2018
#2
+1

Could u go a bit further. I’m a little confused. Or can u solve it.

Thanks!

Guest Aug 23, 2018
edited by Guest  Aug 23, 2018
#3
0

yes can you please solve it?

Thanks, i too am still confused.

Guest Aug 23, 2018
edited by Guest  Aug 23, 2018
#4
+22290
+3

Determine a constant k such that the polynomial

$$\displaystyle P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2)$$

is divisible by x+y+z

$$\begin{array}{|clcll|} \hline \\ P(x, y, z)\text{ is divisible by } x+y+z \\ \\ \hline \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = (x+y+z) \cdot f(x,y,z) \\ \text{Set } x+y+z = 0 \\ \text{so:} \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \cdot f(x,y,z) \\ x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \\ \\ \text{Set arbitrary } x+y+z = 2-1-1= 0 \qquad \text{so } x=2, \quad y=-1, \text{ and } z=-1 \\\\ P(2, -1, -1) = 2^5 + (-1)^5 + (-1)^5 + k\left(2^3+(-1)^3+(-1)^3\right)\left(2^2+(-1)^2+(-1)^2\right) = 0 \\ 32-1-1 + k(8-1-1)(4+1+1) = 0 \\ 30 + k\cdot 6 \cdot 6 = 0 \\ 30 +36k = 0 \\ 36k = -30 \\ k = -\dfrac{30}{36} \\ \mathbf{ k = -\dfrac{5}{6} } \\ \hline \end{array}$$

Example:

$$\begin{array}{|rcll|} \hline x&=& 1 \\ y&=& 2 \\ z &=& 3 \\ x+y+z &=& 6 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline P(1, 2, 3) &=& 1^5 + 2^5 + 3^5 - \dfrac{5}{6}\left(1^3+2^3+3^3\right)\left(1^2+2^2+3^2\right) \\ &=& 1 + 32+ 243 - \dfrac{5}{6}\cdot(1 +8+27)(1+4+9) \\ &=& 276 - \dfrac{5}{6}\cdot 36\cdot 14 \\ &=& 276 - 5\cdot 6\cdot 14 \\ &=& 276 - 420 \\ &=& -144 \\ &=& -24\cdot 6 \quad \text{This is divisible by } x+y+z = 6 \\ \hline \end{array}$$

Aug 23, 2018
#5
+3

THANK YOU SOOOO MUCH!!!!!!!

Guest Aug 23, 2018