Let f(x)= 2x^4 + x^3 - 7x^2 + 4x - 60
Given that f(2i)= 0, use the Complex Conjugate Pair Theorem and Long Division to find all the zeros of f. Hence, factor all f(x)= 2x^4 + x^3 - 7x^2 + 4x - 60 completely.
AAAaannndd the bot gets it wrong yet again!
since 2i and -2i are both roots, (x^2+4) is a factor. So now we have
2x^4 + x^3 - 7x^2 + 4x - 60 = (x^2+4)(2x^2+x-15)
= (x^2+4)(2x-5)(x+3)
Since f(2i)=0, we know that 2i is a root of f(x). By the Complex Conjugate Pair Theorem, we also know that -2i is also a root.
Using long division, we can divide f(x) by the factor (x - 2i)(x + 2i) to obtain the quotient:
2x^2 - 3x - 10
Now, we can factor the quadratic expression to get:
f(x) = (x - 2i)(x + 2i)(2x^2 - 3x - 10)
To find the zeros of the quadratic factor, we can use the quadratic formula:
x = (3 ± √109)/4
Therefore, the complete factorization of f(x) is:
f(x) = (x - 2i)(x + 2i)(x - (3 + √109)/4)(x - (3 - √109)/4)