the quartic equation x^4+2x^3+14x+15=0 has equal root of 1+2i find the other three root
help me show step plz
since the polynomial has real coefficients, complex roots must come in conjugate pairs. SO, it has another root of 1-2i
(x-(1+2i))(x-(1-2i))
= ((x-1)-2i)((x-1)+2i)
= ((x-1)^2 - (2i)^2)
= x^2-2x+5
Now divide the quartic by that, and you will be left with another quadratic, which you can then work out.
sorry but if i do that is the root going to be in same complex number
I just divided the quadratic that Steve found into the origianl quartic and got
x^2 + 4x + 3
which factors to (x+1)(x+3)
So the other two roots are -1 and -3
To find the other three roots of the quartic equation x^4 + 2x^3 + 14x + 15 = 0, given that one of the roots is 1 + 2i, we can use the concept that complex conjugate roots occur in pairs.
Step 1: Use the given root to form a quadratic equation. Since the root is 1 + 2i, the corresponding quadratic equation is (x - (1 + 2i))(x - (1 - 2i)) = 0.
Step 2: Simplify the quadratic equation. Expanding the equation gives: (x - 1 - 2i)(x - 1 + 2i) = 0. Using the difference of squares identity, we have: ((x - 1)^2 - (2i)^2) = 0.
Step 3: Simplify further. Applying the difference of squares results in: (x^2 - 2x + 1 + 4) = 0. This simplifies to: x^2 - 2x + 5 = 0.
Step 4: Find the remaining roots by solving the quadratic equation. We can use the quadratic formula (x = [-b ± √(b^2 - 4ac)] / 2a) to find the other two roots.
For the equation x^2 - 2x + 5 = 0, the values of a, b, and c are:
a = 1, b = -2, and c = 5.
Plugging these values into the quadratic formula, we have:
x = [-(-2) ± √((-2)^2 - 4(1)(5))] / (2(1)).
This simplifies to:
x = [2 ± √(4 - 20)] / 2.
x = [2 ± √(-16)] / 2.
x = [2 ± 4i] / 2.
x = 1 ± 2i.
Hence, the three remaining roots of the quartic equation x^4 + 2x^3 + 14x + 15 = 0 are 1 + 2i, 1 - 2i, and the complex conjugates of 1 ± 2i, which are also 1 ± 2i.