Question 275827: Solve for x.
(x+1)(x+3)(x+5)(x+7)=-15
I've worked this numerous times and get lost every time. Please explain how to find x.
Found 3 solutions by mananth, ikleyn, n2:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! (x+1)(x+3)(x+5)(x+7)=-15
Rearrange the terms
(x+1)x+7)(x+3)(x+5)=-15
(x^2+8x++7)(x^2+8x+15)=-15
Substitute x^2+8x with a
(a+7)(a+15)=-15
a^2+22a +120=-15
a^2+22a+120=0
a^2+12a+10a+120=0
a(a+12)+10(a+12)=0
(a+10(a+12)=0
REPLACE THE VALUE OF A WITH X^2 +8X
and finish
Answer by ikleyn(53727)  (Show Source):
You can put this solution on YOUR website! .
Solve for x.
(x+1)(x+3)(x+5)(x+7)=-15
I've worked this numerous times and get lost every time. Please explain how to find x.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The post by @mananth does not contain and does not give a complete solution.
I came to do this job from the beginning to the end.
Your starting equation is
(x+1)(x+3)(x+5)(x+7) = -15. (1)
Although it leads to the fourth degree polynomial equation, it admits analytical solution
not of transcendent level of complexity.
Let y = x+4. Then
x+3 = y-1, x+5 = y+1, x+1 = y-3, x+7 = y+3.
Therefore, equation (1) can be equivalently written in this form
(y-3)*(y-1)*(y+1)*(y+3 = -15. (2)
Regrouping the terms, we get
(y^2-9)*(y^2-1) = -15,
y^4 - 9y^2 - y^2 + 9 = -15,
y^4 - 10y^2 +24 = 0,
(y^2-6)*(y^2-4) = 0.
Therefore, four routs of equation (2) are
 =  ,  =  ,  = -2,  = 2.
From here, the solutions to equation (1) are
 =  ,  =  ,  = -2-4 = -6,  = 2-4 = -2.
ANSWER. The solutions to the given equation are = , = , = -2-4 = -6, = 2-4 = -2.
Solved.
Answer by n2(76)  (Show Source):
You can put this solution on YOUR website! .
Equations/275827: Solve for x.
(x+1)(x+3)(x+5)(x+7)=-15
I've worked this numerous times and get lost every time. Please explain how to find x.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Your starting equation is
(x+1)(x+3)(x+5)(x+7) = -15. (1)
Although it leads to the fourth degree polynomial equation, it admits analytical solution
not of transcendent level of complexity.
Let y = x+4. Then
x+3 = y-1, x+5 = y+1, x+1 = y-3, x+7 = y+3.
Therefore, equation (1) can be equivalently written in this form
(y-3)*(y-1)*(y+1)*(y+3 = -15. (2)
Regrouping the terms, we get
(y^2-9)*(y^2-1) = -15,
y^4 - 9y^2 - y^2 + 9 = -15,
y^4 - 10y^2 +24 = 0,
(y^2-6)*(y^2-4) = 0.
Therefore, four routs of equation (2) are
= , = , = -2, = 2.
From here, the solutions to equation (1) are
= , = , = -2-4 = -6, = 2-4 = -2.
ANSWER. The solutions to the given equation are = , = , = -2-4 = -6, = 2-4 = -2.
Solved.
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