Respuesta :

Answer:

your answer can be either

Step-by-step explanation:

x = 5

x = 3

 x=  0.0000 - 2.0000 i  

 x=  0.0000 + 2.0000 i  

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 ((((x4)-(8•(x3)))+19x2)-32x)+60  = 0  

Step  2  :

Equation at the end of step  2  :

 ((((x4) -  23x3) +  19x2) -  32x) +  60  = 0  

Step  3  :

Polynomial Roots Calculator :

3.1    Find roots (zeroes) of :       F(x) = x4-8x3+19x2-32x+60

Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  60.  

The factor(s) are:  

of the Leading Coefficient :  1

of the Trailing Constant :  1 ,2 ,3 ,4 ,5 ,6 ,10 ,12 ,15 ,20 , etc  

znk

Answer:

x = ½(1 + √61)

Step-by-step explanation:

The general formula for a fourth-degree polynomial is

f(x) = ax⁴ + bx³ + cx² + dx + e

Your polynomial is  

f(x) = x⁴ - x³ - 19x² + 4x + 60 = 0

a = 1; e = 60

(a) Try to find some rational roots

According to the rational root theorem, the rational roots are

Factors of e/Factors of a

Factors of e = ±1, ±2, ±3,±4, ± 5, ±6, ±10, ±12, ±15, ±20, ±30, ±60.

Factors of a = ±1

Potential roots are x = ±1, ±2, ±3,±4, ± 5, ±6, ±10, ±12, ±15, ±20, ±30, ±60

Now, it's a matter of trial and error to find a zero, and that's a lot of roots.

Rather than go through them all, I will use just the ones that work.

Let's try x = 2 by synthetic division.

2|1   -1  -19      4   60

 |     2    2   -34  -60

  1    1  -17   -30      0

So, x = 2 is a zero, and

(x⁴ - x³ - 19x² + 4x + 60)/(x - 2) = (x³ + x² -17x - 30)(x - 2)

(b) Solve the cubic equation

x³ + x² -17x - 30 = 0

Try x = -2 by synthetic division

-2|1   1    -17  -30  

  |   -2      2   30

   1  -1    -15     0

So, x = -2 is a zero, and x³ + x² -17x - 30 = (x² - x - 15)(x + 2) and

(x⁴ - x³ - 19x² + 4x + 60) = (x² - x - 15)(x - 2)(x + 2)

(c) Solve the quadratic equation

x² - x - 15 = 0

a = 1, b = -1, c = -15

x = [-b ± √(b² - 4ac)]/(2a) = (-b ± √D)/(2a)

D = b²- 4ac = (-1)² - 4×1×(-15) = 1 + 60 = 61

x = (-1 - √61)/(2×1)     x = (-1+ √61)/(2×1)

x = ½(1 - √61)            x = ½(1 + √61)

The four roots are x = ½(1 - √61), x = -2, x = 2, x = ½(1 + √61).

The largest root is x = ½(1 + √61).

The Figure below shows the graph of your polynomial with all the zeroes.

Ver imagen znk

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