If \(x+1\) is a tactor of \(2 x^{3}+3 x^{2}+k x+4\), tind the value of \(k . .\)
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Correct Answer: Option B
Explanation:
Let \(p(x)=\) of \(2 x^{3}+3 x^{2}+k x+4\). Since \(x+1\) is a factor, by factor theorem, \(p(-1)=0\)
\begin{aligned}
&\therefore p(-1)=2(-1)^{3}+3(-1)^{3}+k(-1)+4=0 \\
&=>-2+3-k+4=0 \\
&=5-k=0 \\
&=>k=5
\end{aligned}
Let \(p(x)=\) of \(2 x^{3}+3 x^{2}+k x+4\). Since \(x+1\) is a factor, by factor theorem, \(p(-1)=0\)
\begin{aligned}
&\therefore p(-1)=2(-1)^{3}+3(-1)^{3}+k(-1)+4=0 \\
&=>-2+3-k+4=0 \\
&=5-k=0 \\
&=>k=5
\end{aligned}