Question 156591
Solve for x:
1) {{{x^4 = 5x^2}}} Divide both sides by {{{x^2}}}
{{{x^4-5x^2 = 0}}} Factor out an {{{x^2}}}
{{{x^2(x^2-5) = 0}}} Apply the zero product rule: If a*b = 0, then either a = 0 or b = 0 or both.
{{{x^2 = 0}}} or {{{x^2-5 = 0}}}
If {{{x^2 = 0}}} then {{{x = 0}}} and {{{x = 0}}}
If {{{x^2-5 = 0}}} then {{{x^2 = 5}}} Take the square root of both sides to get:
{{{x = sqrt(5)}}} or {{{x = -sqrt(5)}}}
Now because this is a fourth degree polynomial, you can expect four solutions, some of which may be identical.
The solutions are:
{{{x = 0}}}
{{{x = 0}}}
{{{x = sqrt(5)}}}
{{{x = -sqrt(5)}}}
--------------------------------
2) {{{x^4+18x^3+8x+144 = 0}}} This polynomial can be factored into:
{{{(x^3+8)(x+18) = 0}}} and the first factor can be further factored as it is the sum of two cubes {{{(x)^3}}} and {{{2^3}}}, so we get:
{{{(x+2)(x^2-2x+4)(x+18) = 0}}} Now, applying the zero product rule that was used in the first problem, you will set each of these three factors equal to zero.
{{{x+2 = 0}}} Here, {{{x = -2}}}
{{{x^2-2x+4 = 0}}} Here, we'll have to use the quadratic formula to solve for x: {{{x = (-b+-sqrt(b^2-4ac))/2a}}} and a = 1, b = -2, and c = 4, so...
{{{x = (-(-2)+-sqrt((-2)^2-4(1)(4)))/2(1)}}} Simplifying this, we get:
{{{x = (2+-sqrt(4-16))/2}}} 
{{{x = (2+-sqrt(-12))/2}}}
{{{x = (2+-sqrt(4(-3)))/2}}}
{{{x = (2+2*sqrt(-3))/2}}} or {{{x = (2-2*sqrt(-3))/2}}} Simplifying this:
{{{x = 1+sqrt(3)i}}} or {{{x = 1-sqrt(3)i}}}
...and for the last one:
{{{x + 18 = 0}}}
{{{x = -18}}}
So, as in the first problem, because you have a fourth degree polynomial, you can expect four solutions, and they are:
{{{x = -2}}}
{{{x = 1+sqrt(3)i}}}
{{{x = 1-sqrt(3)i}}}
{{{x = -18}}}
-----------------
3) {{{x^3+19x^2-7x-133 = 0}}} You can solve this by "factoring by grouping"
{{{(x^3+19x)-(7x+133) = 0}}} Now factor each of the two terms:
{{{x^2(x+19)-7(x+19) = 0}}} Factor out the common factor of {{{x+19}}}
{{{(x+19)(x^2-7) = 0}}} Apply the zero product rule.
{{{x+19 = 0}}} or {{{x^2-7 = 0}}} so...
{{{x = -19}}} or {{{x^2 = 7}}} 
{{{x^2 = 7}}} Take the square root of both sides.
{{{x = sqrt(7)}}} or {{{x = -sqrt(7)}}}
The three solutions are:
{{{x = -19}}}
{{{x = sqrt(7)}}}
{{{x = -sqrt(7)}}}
----------------
I'll get you started on the last one but see if you can finish it up yourself.
4) {{{x^6-13x^5-7x = -91}}} Add 91 to both sides.
{{{x^6-13x^5-7x-91 = 0}}} Factor by grouping.
{{{(x^6-13x^5)-(7x-91) = 0}}} Factor an {{{x^5}}} from the first term and  a 7 from the second term.
{{{x^5(x-13)-7(x-13) = 0}}} Now factor the common factor {{{(x-13)}}}
{{{(x-13)(x^5-7) = 0}}} Apply the zero product rule.
{{{x-13 = 0}}} or {{{x^5-7 = 0}}}
Can you finish this?