How do I solve polynomial equation by finding all complex roots?
The problem is:
factor each expression and find all complex roots of x^3+64
I got as far as x^3+64=(x+4)(x^2-4x+16)
Now how do i find the roots of (x+4) and the roots of (x^2-4x+16)
Thanks
the root of x+4=0 is x=-2
the roots of the next can be solved by the quadratic equation..
x=(4+-sqrt(16-64))/2
that gives you three roots.
When a trinomial is factored as (x + m)(x + n), what is the sum of m and n?
A. The coefficient of the x-term of the trinomial
B. The constant term of the trinomial
C. The degree of the trinomial
D. The coefficient of the x2-term of the trinomial
To find all the complex roots of a polynomial equation, you can use the factored form of the polynomial. In your case, you have factored the expression x^3+64 into (x+4)(x^2-4x+16).
To find the roots of (x+4), you set it equal to zero and solve for x:
x + 4 = 0
x = -4
So, the root of (x+4) is -4.
To find the roots of (x^2-4x+16), you can use the quadratic formula, since it is quadratic. The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is x^2-4x+16 = 0, so a = 1, b = -4, and c = 16. Plugging these values into the quadratic formula:
x = (-(-4) ± √((-4)^2 - 4(1)(16))) / (2(1))
x = (4 ± √(16 - 64)) / (2)
x = (4 ± √(-48)) / (2)
x = (4 ± √(48)i) / (2)
x = (4 ± 4√(3)i) / (2)
x = 2 ± 2√(3)i
Therefore, the roots of (x^2-4x+16) are:
x = 2 + 2√(3)i
x = 2 - 2√(3)i
So, the complex roots of x^3+64 are -4, 2 + 2√(3)i, and 2 - 2√(3)i.