Question

Using factor theorem, find the solution set of the equation $\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& 2x& 5x^2\end{array}\right|=0$

Answer 1

$\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& 2x& 5x^2\end{array}\right|=0$

$\Rightarrow 1((-2\times 5x^2)-5\left(2x\right))-4(5x^2-5)+20(2x+2)=0$

$\Rightarrow -10x^2-10x-20x^2+20+40x+40=0$

$\Rightarrow -30x^2+30x+60=0$

$\Rightarrow 30x^2-30x-60=0$

$\Rightarrow 30(x^2-x-2)=0$

$\Rightarrow x^2-x-2=0$

If we put $x=-1$ in the equation

We get,

${(-1)}^2-(-1)-2\Rightarrow 1+1-2=0$

$\therefore (-1)$ is a root

$\therefore (x+1)$ is a factor of $x^2-x-2$

$(x-2)(x+1))\overline{x^2-x-2}{x}^2\pm x--\_\_\_\_\_\_\_\_\_\_\_\_\_-2x-2-2x-2\_\_\_\_\_\_\_\_\_\_\_\_\_0$

$\therefore x^2-x-2=(x+1)(x-2)=0$

$\therefore x=-1$ or $x=2$