The Shorter and Easier Way
We see that there are 2 #(-2y-3)# terms in the equation. This might signal something useful. So, let's temporarily replace #(-2y-3)# with #x#. Now, let's rewrite the equation:
#x^3+25x=0#
Using the Distributive Property, we can factor out an #x# from the left side:
#x*(x^2+25)=0#
Looking at our new equation, we can see that the two terms' products is #0#. This means that either #x# is #0# or #x^2+25# is #0#. Let's start with the case of #x=0# and solve for #y#. However, before we start, remember that #x# is actually #(-2y-3)#:
#x=0#
#-2y-3=0#
#-2y=3#
#y=-3/2#
Besides this single answer of #y=-3/2#, there are two more roots (solutions) which involve complex numbers. To obtain these two other solutions, we have to look at the case where #x^2+25=0#.
#x^2+25=0#
#x^2=-25#
#x=sqrt-25#
#x=sqrt-1*sqrt25#
#x=+-5i#
The final statement shows that #x# can be either positive #5i# of negative #5i#. We'll start by working with positive #5i#. And, because we let #x=(-2y-3)#, we can substitute it in and find the second value for #y#:
#-2y-3=5i#
#-2y=5i+3#
#y=(5i+3)/-2#
#y=(5i)/(-2)+3/(-2)#
#y=(-5/2)i-3/2#
#y=-3/2-(5/2)i#
The third solution for #y# involves seeing #x# as equal to #-5i#:
#-2y-3=-5i#
#-2y=-5i+3#
#y=(-5i+3)/-2#
#y=(-5i)/-2+3/-2#
#y=-3/2+(5/2)i#
The Long and Hard Way
Let's start with the original problem:
#(-2y-3)^3+25(-2y-3)=0#
Let's expand everything out and use the distributive property to simplify things a bit:
#(-2y-3)(-2y-3)(-2y-3)+(25)(-2y)-(25)(3)=0#
#(-2y-3)(4y^2+6y+6y+9)+(-50y)-75=0#
#(4y^2)(-2y-3)+(6y)(-2y-3)+(6y)(-2y-3)+(9)(-2y-3)-50y-75=0#
#(-8y^3-12y^2)+(-12y^2-18y)+(-12y^2-18y)+(-18y-27)-50y-75=0#
Now let's break open the parentheses:
#-8y^3-12y^2-12y^2-18y-12y^2-18y-18y-27-50y-75=0#
And now, let's start rearranging like terms together and simplifying them:
#-8y^3-12y^2-12y^2-12y^2-18y-18y-18y-50y-27-75=0#
#-8y^3-36y^2-104y-102=0#
At this point, we can use a cubic calculator to calculate the value of #y#:
#y=-1.5#
It's important that two other solutions also work perfectly fine, not just #y=-1.5#. The two other solutions are: #y=-3/2+5/2i# and #y=-3/2-5/2i#.
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Checking:
If we were to check whether or not #y# is indeed #-1.5#, we would substitute #y=-1.5# back into the original equation:
#[-2(-1.5)-3]^3+25[-2(-1.5)-3]=0#
#(3-3)^3+25(3-3)=0#
#0^3+25(0)=0#
#0+0=0#
#0=0#
