Question 1189239
Given that {{{ x^2-3x+2 }}} is a factor of {{{ x^4 + kx^3 - 10x^2 - 20x+24 }}} evaluate the sum of the four roots of the equation {{{ x^4 + kx^3 - 10x^2 - 20x+24 = 0 }}} 
<pre>When factored, {{{x^2 - 3x + 2}}} = (x - 1)(x - 2).
As {{{x^2 - 3x + 2}}} is a factor of {{{x^4 + kx^3 - 10x^2 - 20x + 24}}}, x - 1 and x - 2 are also factors of {{{x^4 + kx^3 - 10x^2 - 20x + 24}}}, which 
means that x - 1 = 0, or x = 1, and x - 2 = 0, or x = 2. So, 2 of the roots of {{{x^4 + kx^3 - 10x^2 - 20x + 24}}} are 1 and 2.
Using either root, and the RATIONAL ROOT THEOREM, we find that k = 5.

The equation {{{x^4 + kx^3 - 10x^2 - 20x + 24}}} now becomes: {{{x^4 + 5x^3 - 10x^2 - 20x + 24}}}, and when POLYNOMIAL LONG-DIVISION and its
factor, {{{x^2 - 3x + 2}}} are used, the other factor of the polynomial, {{{x^2 + 8x + 12}}} is derived. 
And, when {{{x^2 + 8x + 12}}} is factored, its roots, from its factors x + 6 and x + 2, are - 6 and - 2.
We now have roots: 1, 2, - 6, and - 2. 
Therefore, the sum of the roots of {{{x^4 + kx^3 - 10x^2 - 20x + 24}}} or {{{x^4 + 5x^3 - 10x^2 - 20x + 24}}} = 1 + 2 + (- 6) + (- 2) = <font color = red><font size = 4><b>- 5.</font></font></b>