Question

Compute the greatest root of x^4-x^3-19x^2+4x+60=0. Express your answer in simplified radical form.

Answer 1

your answer can be either

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

Answer 2

x = ½(1 + √61)

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.