SOLUTION: Factors the polynomial P(x).then solve the equation P(x)=0. 1.P(x)=x^3+4x^2+x-6 2.P(x)=x^3-6x^2-x+6 3.P(x)=x^3-x^2-x+1 4.P(x)=2x^3-3x^2-3x+2 5.P(x)=x^4-x^3-19x^2+49x-30

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Factors the polynomial P(x).then solve the equation P(x)=0. 1.P(x)=x^3+4x^2+x-6 2.P(x)=x^3-6x^2-x+6 3.P(x)=x^3-x^2-x+1 4.P(x)=2x^3-3x^2-3x+2 5.P(x)=x^4-x^3-19x^2+49x-30      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 985474: Factors the polynomial P(x).then solve the equation P(x)=0.
1.P(x)=x^3+4x^2+x-6
2.P(x)=x^3-6x^2-x+6
3.P(x)=x^3-x^2-x+1
4.P(x)=2x^3-3x^2-3x+2
5.P(x)=x^4-x^3-19x^2+49x-30
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
5.

P%28x%29=x%5E4-x%5E3-19x%5E2%2B49x-30=0

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

Try 2

1 | 1  -1 -19  49 -30
  |     1   0 -19  30
    1   0 -19  30   0

So we have factored P(x) as 

P%28x%29=%28x-1%29%28x%5E3%2B0x%5E2-19x%2B30%29=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
  |  2   4 -30
    1   2 -15   0

So we have factored P(x) further as as

P%28x%29=%28x-1%29%28x-2%29%28x%5E2%2B2x-15%29=0%29

Now just factor the trinomial:

P%28x%29=%28x-1%29%28x-2%29%28x%2B5%29%28x-3%29=0%29

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

Edwin