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]
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