SOLUTION: 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 }}}

Notice that x^2-3x+2 = (x-2)*(x-1).


We are given that the polynomial  x^4 + kx^3 - 10x^2 - 20x+24  is divisible by the polynomial  x^2-3x+2 .


Hence, the polynomial  x^4 + kx^3 - 10x^2 - 20x+24  is divisible by  (x-1).


It means (the Remainder theorem) that the number x= 1 is the root of the polynomial  x^4 + kx^3 - 10x^2 - 20x+24.


So, we substitute x= 1 into this polynomial, and we get this equation for "k"

    1^4 + k*1^3 - 10*1^2 - 20*1 + 24 = 0,

or

    1 + k - 10 - 20 + 24 = 0,

    k = 5.


Now use the Vieta's theorem: the sum of the roots of the polynomial  x^4 + kx^3 - 10x^2 - 20x + 24  is equal 
to the coefficient at x^3 with the opposite sign.


It gives that the sum of the roots of the polynomial  x^4 + kx^3 - 10x^2 - 20x+24   is equal to -k, i.e. -5.    ANSWER

When factored,  = (x - 1)(x - 2).

As  is a factor of , x - 1 and x - 2 are also factors of , which 

means that x - 1 = 0, or x = 1, and x - 2 = 0, or x = 2. So, 2 of the roots of  are 1 and 2.

Using either root, and the RATIONAL ROOT THEOREM, we find that k = 5.



The equation  now becomes: , and when POLYNOMIAL LONG-DIVISION and its

factor,  are used, the other factor of the polynomial,  is derived. 

And, when  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  or  = 1 + 2 + (- 6) + (- 2) = - 5.