Question 1159546
The function f is given by f(x)=x^3+2x^2+ax-8 where a is constant.when f(x) is  divided by (x-2) the remainder is -6.
Show that (x+1) is a factor of f(x)
<pre>Factor: x - 2, so we get: x - 2 = 0___x = 2 <==== NON-ROOT
{{{matrix(7,3, f(x), "=", x^3 + 2x^2 + ax - 8,
f(2), "=", 2^3 + 2(2)^2 + a(2) - 8,
- 6, "=", 8 + 8 + 2a - 8,
- 6 - 8, "=", 2a,
- 14, "=", 2a,
(- 14)/2, "=", a,
- 7, "=", a)}}}

Factor: x + 1, so we get: x + 1 = 0___x = - 1 <==== One ROOT
{{{matrix(4,3, f(x), "=", x^3 + 2x^2 - 7x - 8,
f(- 1), "=", (- 1)^3 + 2(- 1)^2 - 7(- 1) - 8,
0, "=", - 1 + 2 + 7 - 8,
0, "=", 0)}}}
As the above proves to be TRUE (0 = 0), x + 1 is DEFINITELY a factor of {{{matrix(1,3, f(x), "=", x^3 + 2x^2 - 7x - 8)}}}
I hope for your SANITY, you won't even look at the UNNECESSARY COMPLEX method presented by the other person!
Plus, the majority of what she has stated is WRONG, leading to WRONG answers, as USUAL!!