Asked by Paul
Find a polynomial of degree 3 with roots 3, 4 - i.
Which one is it?
A.x3 - 10x + 15
B.x3 + 15x2 + 4x - 50
C.x3 - 11x2 + 41x - 51
D.x3 + 24x2 + 3x - 12
Which one is it?
A.x3 - 10x + 15
B.x3 + 15x2 + 4x - 50
C.x3 - 11x2 + 41x - 51
D.x3 + 24x2 + 3x - 12
Answers
Answered by
bobpursley
Well, 3 is a root of C, and the others are not.
See if Google calculator can do the other root; paste this into the google search engine window:
(4-i)^3-11(4-i)^2 +41(4-i)-51
See if Google calculator can do the other root; paste this into the google search engine window:
(4-i)^3-11(4-i)^2 +41(4-i)-51
Answered by
drwls
They all are degree 3.
See if x=3 works for each choice. It does not work for A, B or D.
If one of the choices is correct, it must be (C)
If you want to see if 4-i is a root, long-divide polynomial (C) by (x-3) and see what the roots are for the remaining quadratic, which is
x^2 -8x +17 = 0.
x = (1/2) [8 +/- sqrt(-4)]
= 4 +/- i
See if x=3 works for each choice. It does not work for A, B or D.
If one of the choices is correct, it must be (C)
If you want to see if 4-i is a root, long-divide polynomial (C) by (x-3) and see what the roots are for the remaining quadratic, which is
x^2 -8x +17 = 0.
x = (1/2) [8 +/- sqrt(-4)]
= 4 +/- i
Answered by
Damon
3, 4-i then also 4+i
(x-3)(x-4+i)(x-4-i) = 0
(x-3)(x^2 - 8x + 17)
x^3 - 11 x^2 + 41 x - 51
(x-3)(x-4+i)(x-4-i) = 0
(x-3)(x^2 - 8x + 17)
x^3 - 11 x^2 + 41 x - 51
Answered by
Paul
Hello Damon im not understanding... are you saying the answer is c?
Answered by
drwls
We are all saying the answer is C.
Answered by
Henry
X=3, X-3=0. X=4-i, x-4=-i, Square both
sides:X^2-8X+16=-1,X^2-8X+17=0, Multiply:(X-3)(X^2-8X+17)=0
X^3-11X^2+41X-51=0.
Yes, the correct choice is C.
sides:X^2-8X+16=-1,X^2-8X+17=0, Multiply:(X-3)(X^2-8X+17)=0
X^3-11X^2+41X-51=0.
Yes, the correct choice is C.
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