Really dumb math question...how do I factor a trinomial to find its roots?x^3+x^2+x+1=0
Thanks
Factor a Trinomial?
2008-03-24 4:27 pm
回答 (4)
2008-03-24 4:33 pm
✔ 最佳答案
x^3+x^2+x+1= (x^3+x^2)+(x+1)
= x^2(x+1)+(x+1)
= (x^2+1)(x+1)
So, the roots are: -1, ± i
2008-03-24 11:38 pm
x^3 + x^2 + x + 1 = 0
(x^2 + 1)(x + 1) = 0
x^2 + 1 = 0
x^2 = -1
not a real root
x + 1 = 0
x = -1
(x^2 + 1)(x + 1) = 0
x^2 + 1 = 0
x^2 = -1
not a real root
x + 1 = 0
x = -1
2008-03-24 11:41 pm
By trial and error, (or by graphing) .
-1 is a root.
So, (x+1) is a factor
Divide x^3+x^2+x+1 by x+1
]]]]]]x^2+1
--------------------
x+1)x^3+x^2+x+1
]]]]]]x^3+x^2
--------------------
]]]]]]]]]]]]0x^2+x+1
]]]]]]]]]]]]]]]]]]]]x+1
-----------------------
(x+1)(x^2+1)
There ia only 1 real root -1. The others are complex.
-1 is a root.
So, (x+1) is a factor
Divide x^3+x^2+x+1 by x+1
]]]]]]x^2+1
--------------------
x+1)x^3+x^2+x+1
]]]]]]x^3+x^2
--------------------
]]]]]]]]]]]]0x^2+x+1
]]]]]]]]]]]]]]]]]]]]x+1
-----------------------
(x+1)(x^2+1)
There ia only 1 real root -1. The others are complex.
2008-03-24 11:36 pm
find numbers that will make the left side equal 0. in this case negative 1 works. then u got a factor x+1. then use synthetic divison to find the quadratic root, itwill be of degree 2. then factor that or leave it .
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