[bohemianheart.]

Alrighty, well I'm having a very hard time figuring out these equations. I was sitting at my desk for like two hours figuring out only two questions. But I just don't seem to understand it at all. Here is what the book says for examples:

Example #5: Find a third-degree polynomial equation with rational coefficients that has roots 3 and 1+i.
Step 1: Find the other root using the Imaginary Root Theorem.

Since 1+i is a root, then its complex conjugate 1-i also is a root.

Step 2: Write the factored term of the polynomial using the factor theorem.

Step 3: Multiply the factors.

(X-3)[x^2-x(1-i)-x(1+i)+(1+i)(1-i)]
(x-3)(x^2-x+ix-x-ix+1-i^2)
(x-3)(x^2-2x+2)
x^3-5x^2+8x-6

------------ Well I'm not really sure how to figure out this theorem. & time ran out in class so I don't know what the heck i'm doing. ):

I just need help on two questions:

19. 1 and 3i
20. -5 and 1-i

Please & thank you. I really need help. ): too many things happening at once this week. haha. Thank you so much for who ever helps me. <3333

[Vivskies]

Have you guys learned about:
S1+S2=b/a
S1 x S2= -c/a?

And with S1&S2 as the roots?
And 'a','b' & 'c' being in the equation 0=Ax^2+bx+c?
Have you guys learned about that yet?

I know how to do it like that.

[bohemianheart.]

^

Um.. I never learned that sry. ): My teacher never really taught me how to do this equation.

[ChunJin]

Hint: Roots always has two answers (plus and negative due to their parabolic nature).

The other is easy to identify.

[stevenn go rawrr]

19.
(x-1)[x^2-x(3i)-x(-3i)+(3i)(-3i)]=
(x-1)[x^2-9(-1)] = x^3-x^2+9x-9

20.
(x+5)[x^2-x(1-i)-x(1+i)+(1+i)(1-i)]=
(x+5)[x^2-2x+1-(-1)] = x^3+3x^2-8x+10


look in the back of the book if that's the right answer xD i'm sure they should give you the answer to the odd number questions.

pretty much all you need to know is find the conjugate and know the ____ degree polynomial equation formula (e.i.: ax^3+bx^2+cx+d=0)

[roneddski]

the answers given by stevenn go rawrr r both correct.
actually to check whther u have the correct third-degree polynomial, jus substitute its roots back into the polynomial.
if the results is zero, u r correct.

For example,

19. ans is x^3 - x^2 + 9x - 9

so if 1 is a root, subst it back(meaning x=1), giving 1^3 - 1^2 + 9(1) - 9 = 1 - 1 + 9 - 9 = 0

similarly,

if 3i is a root, subst it back(meaning x=3i), giving (3i)^3 - (3i)^2 + 9(3i) - 9 = -27i + 9 + 27i - 9 = 0

since 3i is a root, its complex conjugate -3i is oso a root. therefore the third-degree polynomial is correct since both substitutions yield zero.

try tis method for question no. 20.