Question

The number of integers $x$ satisfying $-3x^4$ +det $\left[\begin{array}{ccc}1& x& x^2\\ 1& x^2& x^4\\ 1& x^3& x^6\end{array}\right]=0$ is equal to

• A. 1
• B. 2
• C. 5
• D. 8

Answer 1

Answer: (B)

Given, $-3x^4$+det $\left[\begin{array}{ccc}1& x& x^2\\ 1& x^2& x^4\\ 1& x^3& x^6\end{array}\right]=0$
$\Rightarrow x^8+x^5+x^5-x^4-x^7-x^7=3x^4$
$\Rightarrow x^8-2x^7+2x^5-4x^4=0$
$\Rightarrow x^4\left[x^4-2x^3+2x-4\right]=0$
$\Rightarrow x^4\left[x^3\left(x-2\right)+2\left(x-2\right)\right]=0$
$\Rightarrow x^4\left(x^3+2\right)\left(x-2\right)=0$
$\therefore$ $x$ is an integer, so $x=0,2$