x^4-5x=6??

Question

Solve this equation please! I'm trying to learn algebra

Answer 1

x^4-5x=6
x^4 -5x -6 = 0

There's no perfectly straightforward method for finding the factors of a fourth degree polynomial, but the "Rational Root Test" gives you some good candidates when all the coefficients are integers. The list of candidates is made by making fractions from each factor of the constant (-6 in this case) over each of the factors of the highest power (1 in this case).

That gives us [+-1/1, +-2/1, +-3/1, +-6/1]
So trying dividing x^4 -5x -6 by (x-C) where C is each one of the candidates. If you find one that divides evenly, try the remaining candidates against the resulting lower-power polynomial.

(x^4 -5x -6)/(x-1) = x^3+x^2+x-6 with a remainder of -12
So that's no good.

(x^4 -5x -6)/(x+1) = x^3 -x^2 +x -6
So (x+1) works.

(x^3 -x^2 +x -6)/(x-2) = x^2 +x +3
So (x-2) is a factor.

Now that we're down to a quadratic, you can find the remaining roots directly with The Quadratic Equation.

x = (-1 +- sqrt(1 - 12))/2
= (-1+- sqrt(-11))/2
= (-1 +- sqrt(11)i)/2

(x+1)(x-2)(x - ((-1 - sqrt(11)i)/2) (x - ((-1 + sqrt(11)i)/2) = 0
So the two Real roots are -1 and 2.

Let's check those:
(-1)^4 - 5(-1) = 6
1 - -5 = 6 check.

2^4 - 5(2) = 6
16 - 10 = 6 check.

Answer 2

This is a fourth degree polynomial. I would assume you are trying to find the roots of this equation.

x^4 - 5x -6 = 0

the first step is to try and find some common factors:

does x = 1?

1-5-6 = 10

so x is not = 1

is x = 2?

16 -10 -6 = 0
so x = 2 is a factor

Now lets look for some other factors that might work.

does x = -1 work

1+5 - 6 = 0
so x =-1 is a second factor.

We can now divide out (x+1)(x-2) or (x^2 - x - 2) from our equation. The result will give us a quadratic which is easy to factor.

after dividing x^2 - x - 2 into x^4 - 5x -6 we get a nice tailor made equation of x^+x +3.

When we factor that we get

(x + (1+j*sqrt(11))/2) (x + (1 + j*sqrt(11))/2)

The completely factored answer is:

(x + 1)(x - 2)(x + (1+j*sqrt(11))/2) (x + (1 + j*sqrt(11))/2)

Answer 3

try graphing the functions and looking for where y = x^4 - 5x and y = 6 intersect. you find the solutions is x = -1 and x = 2.

in all polynomials, the degree of the polynomial tells you the number of solutions there will be.

a first order polynomial, a line, has one solution.
a quadratic polynomial, i.e., a x^2 +b x + c = d has two solutions
and so on.

so for a quartic function that you have here there are a total of four solutions but two of them are in the complex number plane.

Answer 4

x^4-5x-6=0 factors to (x+1)(x-2)(x²+x+3)=0
so x=-1,2 for the real roots
x= (-1±√11i)/2 for the complex ones.

Answer 5

x^4-5x = 6
x = -1

Answer 6

x^4 - 5x - 6 = 0
(x^2 - 6)(x^2 + 1) = 0
x^2 = 6
X^2 = 1

x = sqrt 6
x = -sqrt 6

if you want imaginary answers too, then
x = i
x = -i

Answer 7

x^4-5x=6

(x^4-5x-6)=(x²+1)(x²-6)

(1,-6)

Answer 8

x^4-5x=6
x^4-5x-6=0
factorize
(x-2)(x+1)(x^2+x+3)
set equal zero
x-2=0
x+1=0
x^2+x+3=0
solve
x-2=0
x=2

x+1=0
x=-1

x^2+x+3=0
x=-1/2+(sqrt(11))/2*i
x=-1/2-(sqrt(11))/2*i


so
x=-1/2+(sqrt(11))/2*i
x=-1/2-(sqrt(11))/2*i
x=-1
x=2