Question

Use the given root to find the solution set of the polynomial x^3+3x^2-8x+10=0,1+i
Answer

{1 + i, 1 - i, -5}

{1 + i, 1 - i, 5}

{1 + i, 1 - i, 5i}

{-5, 5}

Answer 1

other root must be the conjugate \(1-i\)

Answer 2

and the quadratic that gives those two roots is
\[x^2-2x+2\] so you can factor that out and find the other root

Answer 3

huh....

Answer 4

hmmm ok lets go with the first thing i wrote

Answer 5

you are given one root at \(1+i\)
do you know what the other one must be? and why?

Answer 6

no and no

Answer 7

ok the roots of a polynomial with real coefficients come in conjugate pairs, meaning if \(a+bi\) is one root, then so is \(a-bi\) so if \(1+i\) is one root, then so is \(1-i\)

Answer 8

the reason is that they come like that is from the quadratic formula which has
\(x=\frac{-b+\sqrt{b^2-4ac}}{2a}\) and its conjugate \(\frac{-b-\sqrt{b^2-4ac}}{2a}\)
you get the complex number and its conjugate if \(b^2-4ac<0\)

Answer 9

that is how you know that the other root is \(1-i\)
now as for the last root, if you did not understand what i wrote above, we already know that answer C and D are not right, there is only one more root, because the polynomial has degree 3 and we have two of them. also it has to be a real number, because the complex ones come in pairs

Answer 10

so it is either 5 or -5 and i guess you can check which one gives you zero