Step-by-step explanation:
If the polynomial has complex roots, the conjugates of those complex roots must also be part of the zeroes.
Therefore x = -3 - i and x = 3 - i are also zeroes of g(x). However this means that g(x) has 5 solutions when the degree of g(x) is 4. This is impossible, and hence the list is incorrect.
If either of the complex roots (x = -3 + i or x = 3 + i) had a multiplicity of 2, their conjugates will also have a multiplicity of 2, which is impossible as discussed previously.
Therefore x = 1 has a multiplicity of 2.
We have
x⁴ - 8x³ + 23x² - 26x + 10 = (x - 1)²(Ax² + Bx + C).
= (x² - 2x + 1)(Ax² + Bx + C)
= Ax⁴ + (B - 2A)x³ + (A + C - 2B)x² + (B - 2C)x + C
By Comparing Coefficients,
A = 1,
B - 2A = -8,
A + C - 2B = 23,
B - 2C = -26,
C = 10
Solving them we get A = 1, B = -6 and C = 10.
Hence x⁴ - 8x³ + 23x² - 26x + 10
= (x - 1)²(x² - 6x + 10).
Using the Quadratic Formula on x² - 6x + 10,
we find that the correct complex roots are
x = 3 + i and x = 3 - i.
Hence,
the zeroes of g(x) are x = 1, x = 3 + i and x = 3 - i.
