Question

One solution of the equation $x^3-8x^2+16x-3=0$ is $3$.

Find the sum of the remaining solutions.

Short answer

If $3$is a solution of the equation $x^3-8x^2+16x-3=0$then the sum of remaining solutions will be$5$

Step-by-step solution

Step 1. given data

The given equation is

$x^3-8x^2+16x-3=0$

one solution of the equation is$3$

Step 2. factorize the equation

one solution of the equation is $x=3$

divide the expressionlocalid="1646068985153" $x^3-8x^2+16x-3$by $x-3$

localid="1646070020448" $\frac{x-3\begin{array}{c}{x}^2-5x+1\end{array}\begin{array}{c}{x}^3-8x^2+16x-3\\ -(x^3-3x^2)\end{array}}{{\displaystyle \frac{\begin{array}{c}-5x^2+16x-3\\ -(-5x^2+15x)\end{array}}{{\displaystyle \frac{\begin{array}{c}x-3\\ -(x-3)\end{array}}{0}}}}}$

so the equation can be factorized aslocalid="1646070164973" $(x-3)(x^2-5x+1)=0$

Step 3. Apply the quadratic formula

use the quadratic formula to find the solution of$(x^2-5x+1)$

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-5)\pm \sqrt{{(-5)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)} \\ x=\frac{5\pm \sqrt{21}}{2} \\ x=\frac{5+\sqrt{21}}{2},\frac{5-\sqrt{21}}{2} \end{gathered}$$

Step 4. the sum of solutions

Sum of solutions of the equation except $x=3$

$$\begin{gathered} \frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=\frac{5+\sqrt{21}+5-\sqrt{21}}{2} \\ =\frac{10}{2} \\ =5 \end{gathered}$$