SOLUTION: If f(x)= 2x^3-x^2+3x+10 and g(x)= x^3+3x^2+2x+4, determine when f(x)>g(x) using a factor table strategy.

Algebra ->  Inequalities -> SOLUTION: If f(x)= 2x^3-x^2+3x+10 and g(x)= x^3+3x^2+2x+4, determine when f(x)>g(x) using a factor table strategy.      Log On


   



Question 1060642: If f(x)= 2x^3-x^2+3x+10 and g(x)= x^3+3x^2+2x+4, determine when f(x)>g(x) using a factor table strategy.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
If f(x)= 2x^3-x^2+3x+10 and g(x)= x^3+3x^2+2x+4, determine
when f(x)>g(x) using a factor table strategy.

I don't know about "factor table strategy", but
here is the way I'd go about solving the problem:

%22f%28x%29%22%22%22%3E%22%22%22g%28x%29%22

2x%5E3-x%5E2%2B3x%2B10%22%22%3E%22%22x%5E3%2B3x%5E2%2B2x%2B4

Get 0 on the right:

x%5E3-4x%5E2%2Bx%2B6%22%22%3E%22%220

We find the critical values, which are the zeros of x%5E3-4x%5E2%2Bx%2B6

Use synthetic division, trying the factors of 6, which are ±1,±2,±3,±6 

1 | 1 -4  1  6
  |    1 -3 -2
    1 -3 -2  4

Remainder isn't 0.  Try -1

-1 | 1 -4  1  6
   |   -1  5 -6
     1 -5  6  0

So x%5E3-4x%5E2%2Bx%2B6 factors as

(x+1)(x2-5x+6) 

and it factors completely as

(x+1)(x-2)(x-3) > 0

So the critical numbers are -1,2, and 3

Intervals to test (between and beyond those):

1.  %28matrix%281%2C3%2C-infinity%2C%22%2C%22%2C-1%29%29

Test x=-2, (-2+1)(-2-2)(-2-3) > 0
                 (-1)(-4)(-5) > 0
                          -20 > 0
FALSE!

2.  %28matrix%281%2C3%2C-1%2C%22%2C%22%2C2%29%29

Test 0, (0+1)(0-2)(0-3) > 0
            (1)(-2)(-3) > 0
                      6 > 0
TRUE!

3. %28matrix%281%2C3%2C2%2C%22%2C%22%2C3%29%29

Test 2.5, (2.5+1)(2.5-2)(2.5-3) > 0
               (3.5)(0.5)(-0.5) > 0
                         -0.875 > 0
FALSE!

%28matrix%281%2C3%2C3%2C%22%2C%22%2Cinfinity%29%29

Test 4, (4+1)(4-2)(4-3) > 0
              (5)(2)(1) > 0
                     10 > 0
TRUE!

So f(x) > g(x) when  -1 < x < 2, and when x > 3

Edwin