Question

3. For the polynomial g(x) = x^4 - 8x^3 + 23^2 - 26x + 10, you are told that the zeros include x=1, x = -3 + i, and x = 3+i.

Answer 1

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.