(x+1) (x+2) (x+3)(x+4) =120 solve this eq
Answers
Answer:
120=(x+1)(x+2)(x+3)(x+4)=((x+1)(x+4))((x+2)(x+3))
=((x2+5x+5)−1)((x2+5x+5)+1)…(1)
=(x2+5x+5)2−1,
we get x2+5x+5=±11.
Thus x2+5x−6=0 or x2+5x+16=0. The first of these give x=1 or x=−6. The second of these give x=12(−5±−39−−−−√).
It is easy to verify that x=1 and x=−6 are both solutions; in fact, 120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2).
That the second pair of complex conjugates is also a solution can be best verified by replacing x2+5x by −16 in eqn. (1), say.
There are two real solutions, x=1 and x=−6, and two non-real solutions x=12(−5±−39−−−−√). ■
We are trying to find the roots of y=(x+1)(x+2)(x+3)(x+4)−120
If we want to use the rational root theorem, all we really care about if the leading coefficient and the constant:
y=x4+⋯+4!−120=x4+⋯−96
So all rational roots will be contained in:
1,2,3,4,6,8,12,16,24,32,48,96,-96,-48,-32,-24,-16,-12,-8,-6,-4,-3,-2,-1
(If I made an error, please tell me, but all rational roots are in there)
After testing all of them, x=−6 and x=1 are roots.
Expanding (x+6)(x−1) to x2+5x−6 and expanding (x+1)(x+2)(x+3)(x+4)−120 to factor out (x+6)(x−1) with polynomial long division:
y=(x+6)(x−1)(x2+5x+16)
Solving x2+5x+16 using the quadratic formula gives the rest of the solutions, so the solutions to (x+1)(x+2)(x+3)(x+4)=120 are:
x=−6
x=1
x=−52+39√2i
x=−52−39√2i
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(x+1)(x+2)(x+3)(x+4)=120
⟹(x+1)(x+4)(x+2)(x+3)=120
⟹(x2+5x+4)(x2+5x+6)=120
Letx2+5x=y
(y+4)(y+6)=120
⟹y2+10y+24=120
⟹y2+10y−96=0
⟹y2+16x−6x−96=0
⟹y(y+16)−6(y+16)=0
(y+16)(y−6)=0
⟹y=−16andy=6
∴x2+5x=6⟹x2+5x−6=0
x2+6x−x−6=0
x(x+6)−1(x+6)=0
(x+6)(x−1)=0
⟹x=−6andx=1real solutions
also
x2+5x=−16⟹x2+5x+16=0
x=−5±25−64−−−−−−√2
⟹x=−5±−39−−−−√2
x=−52+39−−√2icomplex solution
and
x=−52−39−−√2icomplex solution