Solve the polynomial \(x^{3}+4 x^{2}+x-6=0\)
A. \(0,-1,2\) B. \(-2,4,5\) C. \(1,-3,-4\) D. \(0,1,2\) Correct Answer: AExplanation\(x^{3}+4 x^{2}+x-6=0\) Let \(P(x)=x^{3}+4 x^{2}+x-6\) Now, \(P(1)=1+4+1-6=0\). Hence, \(x-1\) is a factor of \(P(x)\) by factor theorem.
\begin{aligned} &x^{2}+5 x+6\\ &x-1 \mid \begin{aligned} &x^{3}+4 x^{2}+x-6 \\ &x^{3}-x^{2} \end{aligned}\\ &5 x^{2}+x\\ &5 x^{2}+x\\ &\frac{6 x-6}{0} \end{aligned}
\(x^{2}+5 x+6=x^{2}+2 x+3 x+6\) \(=(x+2+(x+3)\) \(\therefore x^{3}+4 x^{2}+x-6=0\) \(\Rightarrow(x-1)(x+2)(x+3)=0\) \(\Rightarrow\) either \(x-1=0\) or \(x+2=0\) or \(x+=0\) \(\therefore x=-3,-2,1\)
|