Question

Find all the zeros of the equation x^4-6x^2-7x-6=0

Answer 1

Answer:

The zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}$

$-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2$

$=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}$

$=x^3-2x^2-2x-3$

Again factor using the rational root test, we get

$=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)$

Using the zero product rule, we have

$x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Therefore, the zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Answer 2

Answer:

The zeroes of $x^4-6x^2-7x-6=0$   is 3,-2,   $\frac{-1+ i\sqrt{3}}{2}$  , $\frac{-1- i\sqrt{3}}{2}$ ..

Step-by-step explanation:

Given : $x^4-6x^2-7x-6=0$

To Find: Zeroes of equation $x^4-6x^2-7x-6=0$

Solution:

$x^4-6x^2-7x-6=0$

$(x-3)(x+2)(x^2+x+1)=0$

Set x-3 =0

So, x = 3 is the first solution.

Set x+2=0

So, x =-2 is the second solution.

Now Set $x^2+x+1=0$

Using quadratic formula :

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values a = 1 , b=1 , c=1

$x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$

$x=\frac{-1\pm i\sqrt{3}}{2}$

So, $x=\frac{-1+ i\sqrt{3}}{2},\frac{-1- i\sqrt{3}}{2}$

Hence the zeroes of $x^4-6x^2-7x-6=0$   is 3,-2,   $\frac{-1+ i\sqrt{3}}{2}$  , $\frac{-1- i\sqrt{3}}{2}$ ..