Question

If $a{x}^4+b{x}^3+c{x}^2+dx+e=\left|\begin{array}{ccc}{x}^3+3x& x-1& x+3\\ x+1& -2x& x-4\\ x-3& x+4& 3x\end{array}\right|$, then e =

• A. 1
• B. 0
• C. 2
• D. -1

Answer 1

Answer: (B)

0

We have,

$a{x}^4+b{x}^3+c{x}^2+dx+e=\left|\begin{array}{ccc}{x}^3+3x& x-1& x+3\\ x+1& -2x& x-4\\ x-3& x+4& 3x\end{array}\right|$

Substitute $x=0$ both sides,

$\Rightarrow a\left(0\right)+b\left(0\right)+c\left(0\right)+d\left(0\right)+e=\left|\begin{array}{ccc}0& -1& 3\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|$

$\Rightarrow e=0+1(0-12)+3(4-0)=-12+12=0$, expand the determinant along first row.