SOLUTION: My math question is as follows: Determine a polynomial P(x) with real coefficients and of degree 5 for which three of its roots are x=-1; x=1+2i; and x=-3i. I am setting my

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: My math question is as follows: Determine a polynomial P(x) with real coefficients and of degree 5 for which three of its roots are x=-1; x=1+2i; and x=-3i. I am setting my      Log On


   



Question 219501: My math question is as follows: Determine a polynomial P(x) with real coefficients and of degree 5 for which three of its roots are x=-1; x=1+2i; and x=-3i.

I am setting my problem up as follows and wondering if this is correct: P(x) = (x+1)[x+(1+2i)][x+(1-2i)][x+(3i)][x+(-3i)] OR would the last 2 just be x+3i and x+(-3i).
Thanks Lori

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
You are very close. Most of your +'s should be -'s. The general form would be:
P%28x%29+=+%28x+-+r%5B1%5D%29%28x+-+r%5B2%5D%29%28x+-+r%5B3%5D%29... where r%5Bn%5D is a root. We want the minus signs so that when x is one of the roots, the minus makes the factor with that root after the minus sign zero which, in turn, makes the entire product zero.

So your function should be:

Simplifying within the parentheses we get:
P%28x%29+=+%28x+%2B+1%29%28x+-+1+%2B+2i%29%28x+-+1+-+2i%29%28x+-+3i%29%28x+%2B+3i%29

To multiply this out take advantage of the pattern: %28a+%2B+b%29%28a+-+b%29+=+a%5E2+-+b%5E2. The second and third factors fit this pattern with "a" = "x-1" and "b" = 2i. The fourth and fifth factors also fit this pattern more obviously. So we'll start by multiplying the 2nd and 3rd factors and multiplying the 4th and 5th factors. (Remember multiplication is commutative and associative so we can multiply in any order we choose.)
P%28x%29+=+%28x+%2B+1%29%28%28x+-+1%29%5E2+-+%282i%29%5E2%29%28x%5E2+-+%283i%29%5E2%29
Simplifying we get:
P%28x%29+=+%28x+%2B+1%29%28x%5E2+-2x+%2B+1+-+4i%5E2%29%28x%5E2+-+9i%5E2%29
P%28x%29+=+%28x+%2B+1%29%28x%5E2+-2x+%2B+1+-+4%28-1%29%29%28x%5E2+-+9%28-1%29%29
P%28x%29+=+%28x+%2B+1%29%28x%5E2+-2x+%2B+1+%2B+4%29%28x%5E2+%2B+9%29
P%28x%29+=+%28x+%2B+1%29%28x%5E2+-2x+%2B+5%29%28x%5E2+%2B+9%29
At this point the i's have disappeared. We still have some multiplications to go. I think the easiest way is to multiply the 1st and third factors first:
P%28x%29+=+%28x%5E3+%2B+x%5E2+%2B+9x+%2B+9%29%28x%5E2+-2x+%2B5%29
And finally we multiply each term of the first factor by each term of the second factor:

Adding like terms:
P%28x%29+=+x%5E5+-+x%5E4+%2B+12x%5E3+-+4x%5E2+%2B+27x+%2B+45