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egoroff_w [7]
3 years ago
14

Find all zeros of the function

Mathematics
1 answer:
Pavel [41]3 years ago
6 0

Answer:

x=-1,-3,2i and -2i

Step-by-step explanation:

By plugging in x=-1, f(x)=0 and x=-3,f(x)=0

That means x+1 and x+3 are roots of the fourth degree polynomial.

(x+1)(x+3)=x²+4x+3

Divide the polynomial by x²+4x+3 and that gives x²+4.

We now find the zeros of the factors

(x²+4x+3)(x²+4)=0

x=-1,-3,2i and -2i

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Find x when : 3x^3 - 2x +1 = 0
notsponge [240]
Answer:  x = "-1 " ;  <u>and</u>:  "(1 ± √3) / 2" .
____________________________________________
Explanation:
____________________________________________
Given:  3x³ −  2x  + 1 = 0 ;  Find "x" ; 

Factor:  "3x³ −  2x + 1"  ;

3x³ − 2x + 1 = (3x² <span>− 3x + 1)(x + 1) ;

Now, set the equation equal to "0" (zero);

</span> (3x² − 3x + 1)(x + 1) = 0 ;

x = 0  ; when:
__________________________________
(x + 1) = 0 ;  and/or: when:

(3x² − 3x + 1) = 0 ;
__________________________________

Start with "x + 1 = 0" ;

Subtract "1" from each side of the equation ; to isolate "x" on one side of the <span>equation ; and</span> to solve for "x" ; 

              x + 1 − 1 = 0 − 1 ; 

                x = -1 ;

_______________________________
Now, try:  

3x² − 3x + 1 = 0 ;

Since:  "3x² − 3x + 1 " cannot be factored; and since:

  "3x² − 3x + 1 = 0" ; is written in the "quadratic equation format"; that is:

  "ax² + bx + c = 0 ;  a ≠ 0;  We can use the "quadratic equation formula" to solve for "x" ;  

in which "a = 3" ;   
              "b = -3" ;
              "c = 1" ;
     
The quadratic equation formula is:
__________________________________

x = [-b ± √(b²−4ac] / 2a ;

   = ?  (Let us plug in our known values for "a, b, & c"l 

x = { [-(-3] ± [√[(-3²) − (4*3*1)] } / {2* 3} ;

x = [3 ± √(9 − 12)] / 6 

x = [3 ± √(-3 )] / 6 ; 

x = (1 ± √3) / 2 .
<span>__________________________________________________
So, the answers are:  x = " -1 " ;  and:  "</span>(1 ± √3) / 2 " .
__________________________________________________
8 0
3 years ago
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