Two equations, one synthetic division; two roots of function

[Jaganath]

1) Use synthetic division to do the division in the rational function:
x^5+2x^3-x^2+1/x+2.

2) If -1-3i is a root of a polynomial function g(x), find all the roots of the function g(x)=x^3+4x^2+14x+20.

 

[Jaganath]

Re: Two equations, one synthetic division; two roots of func

1) Use synthetic division to do the division in the rational function:
x^5+2x^3-x^2+1/x+2.

2) If -1-3i is a root of a polynomial function g(x), find all the roots of the function g(x)=x^3+4x^2+14x+20.

So work I've done:

1) x+2=0, x=-2.
I've tried using 2 and -2 but the numbers keep getting bigger they don't cancel out to 0. But I think I might have gotten it. I used the synthetic division and came up with x^4-2x^3=6x^2-13x+21 - (51/x+2)

2) 1-3i; Conjugate: 1 + 3i
I did some multiplying and came out to -1 -6i +9 because factoring them together like I was wasn't yielding good results. But I could be wrong.

 

[Deleted member 4993]

2) If -1-3i is a root of a polynomial function g(x), find all the roots of the function g(x)=x^3+4x^2+14x+20.

This is a cubic equation - hence it has three roots.

Given one root is complex (a + ib).

It must have another root - which is conjugate of the above (a-ib)

Then, if the third root is c

g(x) = [x - (a+ib)][x-(a-ib)][x-c] = [(x-a)^2 + b^2][x-c]

x-c = g(x)/[(x-a)^2 + b^2]

Find 'c' by synthetic division or other favorite method.

 

[soroban]

Re: Two equations, one synthetic division; two roots of func

Hello, Jaganath!

1) Use synthetic division to do the division in the rational function:

. . \(\displaystyle \L\frac{x^5\,+\,2x^3\,-\,x^2\,+\,1}{x\,+\,2}\)

Your answer is correct except for the "21", so I assume it's just a typo.

 -2  |  1   0   2  -1   0   1
     |     -2   4 -12  26 -52
     -------------------------
        1  -2   6 -13  26 -51

Answer: .\(\displaystyle \L x^4\,-\,2x^3\,+\,6x^2\,-\,13x\,+\,26\,-\,\frac{51}{x\,+\,2}\)