2X^4-9X^3+13X^2-X-5=0
Given that one root is 2+i, solve the equation.
So far I have it set up like this:
2X^4-9X^3+13X^2-X-5/X^2-4X+5
I don't know how to get the answer, I know that I have to use long division but when I started to try it I kept getting weird answers. Please Help
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To solve the equation, you can use the fact that complex roots occur in conjugate pairs. Since one root is 2+i, the other root must be its conjugate, which is 2-i.
Let's set up the polynomial with the given roots factored out:
(X - (2+i))(X - (2-i)) = 0
To simplify, we can multiply this out:
(X - 2 - i)(X - 2 + i) = 0
Expanding this equation gives:
(X - 2)^2 - (i)^2 = 0
Let's simplify further:
(X - 2)^2 - i^2 = 0
(X - 2)^2 + 1 = 0
Now, let's go back to the original equation:
2X^4 - 9X^3 + 13X^2 - X - 5 = 0
We can rewrite it by factoring out the roots:
(X - 2)^2 + 1 = 0
Comparing this with the original equation, we can see that X^2 - 4X + 5 = (X - 2)^2 + 1.
Now, divide the original equation by X^2 - 4X + 5 using long division:
-2X^2 -4X +7
--------------------
X^2 - 4X + 5 | 2X^4 - 9X^3 + 13X^2 - X - 5
-2X^4 +8X^3 -10X^2
--------------------
-X^3 +3X^2 -X
+X^3 -4X^2 +5X
--------------------
-X^2 +4X -X
+X^2 -4X +5
--------------------
-X +5
After the long division, we can see that the remainder is -X + 5.
So, we have:
(X^2 - 4X + 5)(2X^2 - X - 1) = 0
Setting each factor equal to zero gives two more possible roots:
X^2 - 4X + 5 = 0 ---> Root: X = 2 ± i
2X^2 - X - 1 = 0 ---> Solve this equation to find two more roots
Now, you can solve the quadratic equation 2X^2 - X - 1 = 0 using the quadratic formula or factoring methods to find the remaining two roots.