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Without using a calculator, determine the number of real zeros of the function f(x) = x.3 + 4x.2 + x – 6.

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4 Answers

You can use factor theorem and synthetic division to answer the question.
f(x) = x3 + 4x2 + x - 6
Since f(1) = 0, by factor theorem x-1 is a factor.
Using synthetic division,
1 | 1 4 1 -6
........1 5 6
-----------------
....1 5 6 | 0
So, f(x) = (x-1)(x2+5x+6) = (x-1)(x+2)(x+3)
 
Answer: There are three real zeros: x = 1, -2, -3.
 X^3  + 4 X^2 + X -6
 
       Sum of the roots has to be : X1 + X2 + X3 = -4
                                                 X1 . X2. X3 = -6              write down all factors of 6 : 1,2, 3
 
                                                                                       we see that -3 - 2 +1 =-4
 
                                                                                                         and (-3)( -2) (1) = -6
 
      Therefore the 3 rational roots are: X1 = -3   X2 = -2   X3=1
 
 Polynomial can be factored as:  X^3 + 4 X^2 + X - 6  =  ( X+3) ( X+2) ( X-1)
  This is what you do without calculation:
 
 
Since the function is a polynomial with real coefficients, and is of odd degree, it must have at least one real root.
Using the rational roots theorem all the roots must be plus or minus factors of 6.  As you can see 1 is such a root.
Using synthetic division
1|1  4  1  -6
       1  5   6
   1  5  6   0  that is x2+5x+6.  So the polynomial is (x-1)(x2+5x+6)                                                               IF ALL YOU ARE AFTER IS HOW MANY REAL ROOTS then
52-4×6=1, the discriminant is positive, the other roots are also real.
Dear Queneshia,
 
I hope I'm interpreting this correctly:
 
f(x) = x3 + 4x2 + x - 6 = 0
 
The expression for f(x) cannot, unfortunately, be factored.
 
One solution that suggests itself immediately is x = 1
 
F(1) = (1)3 + 4*(1)2 +1 = 6 (which is true)
 
Are you allowed to graph the function, Queneshia?  Sometimes for expressions that cannot be factored, graphing might be your only option.
 
Let's make a table of x and f(x) values to graph:
 
x          f(x)
 
-4         -10
-3            0
-2            0
-1           -4
 0           -6
 1            0
 2           20
 
Once you graph the function, you can easily see that the zeros are -3, -2 and 1.  There are no others.

Comments

Question:
                  X1 + X2 + X3 = -b/a = -3 - 2 + 1 = -4    ( True)
 
                 Then :
                             X1 . X2 . X3 = d/a = -6, however  (1) ( -3) ( -2) = +6 , doesn't seem to be consistent.
 
Parviz:  (x-a)(x-b)(x-c)=x3-(a+b+c)x2+(ab+ac+bc)x-abc
In our case -(1)(-2)(-3)=-6

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