Asked by philip
the quartic equation x^4+2x^3+14x+15=0 has equal root of 1+2i find the other three root
help me show step plz
help me show step plz
Answers
Answered by
Steve
since the polynomial has real coefficients, complex roots must come in conjugate pairs. SO, it has another root of 1-2i
(x-(1+2i))(x-(1-2i))
= ((x-1)-2i)((x-1)+2i)
= ((x-1)^2 - (2i)^2)
= x^2-2x+5
Now divide the quartic by that, and you will be left with another quadratic, which you can then work out.
(x-(1+2i))(x-(1-2i))
= ((x-1)-2i)((x-1)+2i)
= ((x-1)^2 - (2i)^2)
= x^2-2x+5
Now divide the quartic by that, and you will be left with another quadratic, which you can then work out.
Answered by
philip
sorry but if i do that is the root going to be in same complex number
Answered by
Reiny
I just divided the quadratic that Steve found into the origianl quartic and got
x^2 + 4x + 3
which factors to (x+1)(x+3)
So the other two roots are -1 and -3
x^2 + 4x + 3
which factors to (x+1)(x+3)
So the other two roots are -1 and -3
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