Question

If x=2 is a zero f(x) x^3 -8x^2+25x-26 algebraically find all zeros. Express in complex roots simplest a+bi form.

Answer 1

x = 3 + 2i or x = 3 - 2i

Explanation:

Given the polynomial equation below

`f(x)=x^3-8x^2\text{ + 25x -26}`

From the question provided, you can see that x= 2 is one of the zero of the above polynomial.

Therefore, we can find the other zeros by first finding the equation by x - 2 using long division

From the calculation above, the quotient is given as

`x^2\text{ - 6x + 13}`

The next step is to solve the quadratic equation using the general formula

Recall that, the standard quadratic equation is given below as

`y=ax^2\text{ + bx + c}`

Hence,

a = 1

b = -6

c = 13

`\begin{gathered} \text{General quadratic formula is} \\ x\text{ }\frac{-b\text{ }\pm\sqrt[]{b^2\text{ - 4ac}}}{2a} \\ \end{gathered}`

Substitute the values of a, b, and c into the formula

`\begin{gathered} x\text{ = }\frac{-(-6)\pm\sqrt[]{(-6)^2\text{ - 4 }\cdot\text{ 1 }\cdot\text{ 13}}^{}}{2\cdot\text{ 1}} \\ x\text{ = }\frac{6\text{ }\pm\sqrt[]{36\text{ - 52}}}{2} \\ x\text{ = }\frac{6\pm\sqrt[]{-16}}{2} \\ \text{ }\sqrt[]{-16}\text{ = }\sqrt[]{-1}\cdot\text{ }\sqrt[]{16} \\ \text{where }\sqrt[]{-1}\text{ = i and }\sqrt[]{16}\text{ = 4} \\ \sqrt[]{-16}\text{ = 4i} \\ \text{Therefore, } \\ \text{x = }\frac{6\text{ + 4i}}{2}\text{ or }\frac{6\text{ - 4i}}{2} \\ \text{x = }3\text{ + 2i or x = 3 - 2i} \end{gathered}`

Therefore, the other zero values are, 3 + 2i and 3 - 2i