High School

Find all real and complex solutions. Show your work.

\[
(2x^{4} + 23x^{3} + 60x^{2})(-125x - 500) = 0
\]

Answer :

The real solutions to the equation [tex](2x^4 + 23x^3 + 60x^2)(-125x - 500) = 0 are x = 0, x = -4, and x = -7.[/tex]

To find the real and complex solutions of the equation [tex](2x^4 + 23x^3 + 60x^2)(-125x - 500) = 0,[/tex] we need to solve each factor separately.

Let's start with the first factor, [tex]2x^4 + 23x^3 + 60x^2 = 0:[/tex]

Factor out the common factor [tex]x^2:x^2(2x^2 + 23x + 60) = 0[/tex]

Now, we can solve each factor separately:

[tex]x^2 = 0:x = 0 (real solution)2x^2 + 23x + 60 = 0:[/tex]

We can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:

[tex]x = (-b ± √(b^2 - 4ac))/(2a)[/tex]

Substituting a = 2, b = 23, c = 60 into the formula, we have:

[tex]x = (-23 ± √(23^2 - 4260))/(2*2)x = (-23 ± √(529 - 480))/(4)x = (-23 ± √49)/(4)x = (-23 ± 7)/(4)[/tex]

So, the solutions for this factor are:

[tex]x = (-23 + 7)/4 = -4 (real solution)x = (-23 - 7)/4 = -7 (real solution)[/tex]

Next, let's consider the second factor,[tex]-125x - 500 = 0:-125x = 500x = 500/(-125)x = -4 (real solution)[/tex]

Therefore, the real solutions to the equation[tex](2x^4 + 23x^3 + 60x^2)(-125x - 500) = 0 are x = 0, x = -4, and x = -7.[/tex]

Learn more about equation from

https://brainly.com/question/29174899

#SPJ11