SOLUTION: The Fundamental Theorem of Algebra Use the given zero to find the remaining zeros of each polynomial function. P(x)=x^4-6x^3+71x^2-146x+530; 2+7i x intercepts = 2+7i & 2-7i

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: The Fundamental Theorem of Algebra Use the given zero to find the remaining zeros of each polynomial function. P(x)=x^4-6x^3+71x^2-146x+530; 2+7i x intercepts = 2+7i & 2-7i       Log On


   

Question 134721: The Fundamental Theorem of Algebra
Use the given zero to find the remaining zeros of each polynomial function.
P(x)=x^4-6x^3+71x^2-146x+530; 2+7i
x intercepts = 2+7i & 2-7i
x=2-7i
=x-2+7i=0=(x-(2-7i))=0
(x-(2-7i))(x-(2+7i))=
x^2-x(2+7i)-x(2-7i)+(2+7i)(2-7i)=
x^2-2x-7xi-2x+7xi+53=
x^2-4x+53=
x^4-6x^3+71x^2-146x+530/x^2-4x+53=
x^2-2x+10=
(x^2-4x+53)(x^2-2x+10)
Now I'm stuck. Could really use some help
Thanks so much
Found 3 solutions by jim_thompson5910, vleith, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You have the correct steps.

All that you need to do is solve for x in x%5E2-2x%2B10


Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve x%5E2-2%2Ax%2B10=0 ( notice a=1, b=-2, and c=10)




x+=+%28--2+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A10+%29%29%2F%282%2A1%29 Plug in a=1, b=-2, and c=10



x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A10+%29%29%2F%282%2A1%29 Negate -2 to get 2



x+=+%282+%2B-+sqrt%28+4-4%2A1%2A10+%29%29%2F%282%2A1%29 Square -2 to get 4 (note: remember when you square -2, you must square the negative as well. This is because %28-2%29%5E2=-2%2A-2=4.)



x+=+%282+%2B-+sqrt%28+4%2B-40+%29%29%2F%282%2A1%29 Multiply -4%2A10%2A1 to get -40



x+=+%282+%2B-+sqrt%28+-36+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%282+%2B-+6%2Ai%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%282+%2B-+6%2Ai%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=1+%2B+3%2Ai or x=1+-+3%2Ai



So the remaining zeros are

x=1+%2B+3%2Ai or x=1+-+3%2Ai

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
You've done great work ... and almost all the hard part too.
In order to be a root, the value of
%28x%5E2-4x%2B53%29%28x%5E2-2x%2B10%29 must be 0 at that point
In order for a product to be 0, one or both terms must be zero.
So set %28x%5E2-2x%2B10%29=0 and solve using the quadratic equation
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
You don't need to solve the term x%5E2-4x%2B53%29 since you generated that from the two given root.
Nice work!

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Everything you have done so far is absolutely correct, but you stopped short of saying that:

x%5E4-6x%5E3%2B71x%5E2-146x%2B530=%28x%5E2-4x%2B53%29%28x%5E2-2x%2B10%29=0 => x%5E2-4x%2B53=0 or x%5E2-2x%2B10=0.

You already have the roots for x%5E2-4x%2B53=0 since you were given one of them and used it and its conjugate to derive the trinomial in the first place. Now all you need is the roots of the quotient polynomial that came from division of the quartic by x%5E2-4x%2B53. In other words, just solve

x%5E2-2x%2B10=0

This one adapts readily to completing the square:

x%5E2-2x=-10

x%5E2-2x%2B1=-10%2B1

%28x-1%29%5E2=-9

x-1=3i or x-1=-3i

x=1%2B3i or x=1-3i

In sum, the four roots of x%5E4-6x%5E3%2B71x%5E2-146x%2B530=0 are:

2%2B7i
2-7i
1%2B3i
1-3i