Question

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve. mc017-1.jpg and x = 1 mc017-2.jpg and x = –1 mc017-3.jpg and x = –1 mc017-4.jpg and x = 1

Answer 1

The equation  $x^6+6x^3+5=0$ which is a sixth order equation can be solved easily by making a substitution that turns the equation into an equivalent quadratic equation.

We will make the substitution, $x^3=y$ , to simplify the equation. The two roots of this equation are $x=-1,x=-5^{{1}/{3}}$

$x^6+6x^3+5=(x^3)^2+6x^3+5=y^2+6y+5=0$.

The next step is to solve for $y$ in the equation on the right by factorization. We will look for two factors of 5 that add up to 6 and then use those factors to factorize the quadratic equation.

$y^2+6y+5=0\\(y+1)(y+5)=0\\=>y=-1,y=-5$

The next step is to change back from $y$ to $x$ starting with

$y=-1$.

$x^3=y\\x^3=-1\\=>x=(-1)^{{1}/{3}}=-1$.

The next step is to work out the value of x when  $y=-5$.$x^3=y\\x^3=-5\\=>x=(-5)^{{1}/{3}}=-5^{{1}/{3}}$

Answer 2

We are given equation :$x^6+6x^3+5=0$

$Use\:the\:rational\:root\:theorem$

$\mathrm{Therefore,\:we\:need\:to\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:5}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$\mathrm{Compute\:}\frac{x^6+6x^3+5}{x+1}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^5-x^4+x^3+5x^2-5x+5$

Therefore, final factored form it

$x^6\:+\:6x^3\:+\:5=\left(x+1\right)\left(x^5-x^4+x^3+5x^2-5x+5\right)$

We can't factor it more.

Therefore,

x+1=0.

x=-1.

Therefore, the real solution of the equation would be -1.