Question 985474
<pre>
5.

{{{P(x)=x^4-x^3-19x^2+49x-30=0}}}

Possible rational solutions are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30

Try 2

1 | 1  -1 -19  49 -30
  |<u>     1   0 -19  30</u>
    1   0 -19  30   0

So we have factored P(x) as 

{{{P(x)=(x-1)(x^3+0x^2-19x+30)=0}}}

Possible rational solutions to the second parenthetical expression
set = 0 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30


Try 2

2 | 1   0 -19  30
  |<u>  2   4 -30</u>
    1   2 -15   0

So we have factored P(x) further as as

{{{P(x)=(x-1)(x-2)(x^2+2x-15)=0)}}}

Now just factor the trinomial:

{{{P(x)=(x-1)(x-2)(x+5)(x-3)=0)}}}

x-1=0;  x-2=0;  x+5=0;  x-3=0
  x=1;    x=2     x=-5    x=3

Edwin</pre>