# What are the roots of #(x+1)(2x+3)(2x+5)(x+3) = 945# ?

##### 1 Answer

The roots are:

#x = 2# ,#x = -6# ,#x = -2+-sqrt(59)/2i#

#### Explanation:

Given:

#(x+1)(2x+3)(2x+5)(x+3) = 945#

Notice that:

#(x+1)# and#(x+3)# differ by#2#

#(2x+3)# and#(2x+5)# differ by#2#

What are the factors of

The prime factorisation is:

#945 = 3^3*5*7#

So we have:

#945 = 3*5*7*9 = (color(red)(2)+1) * (color(red)(2)+3) * (2*color(red)(2)+3) * (2*color(red)(2)+5)#

So one solution is

What about other solutions?

Let

Then:

#0 = (x+1)(2x+3)(2x+5)(x+3) - 945#

#color(white)(0) = (t-1)(2t-1)(2t+1)(t+1) - 945#

#color(white)(0) = (t^2-1)(4t^2-1) - 945#

#color(white)(0) = 4t^4-5t^2-944#

We know that

#0 = 4t^4-5t^2-944 = (t^2-16)(4t^2+59)#

Hence the other two roots are

So the roots of our transformed quartic are:

#t = +-4# and#t = +-sqrt(59)/2i#

Then

#x = 2# ,#x = -6# ,#x = -2+-sqrt(59)/2i#