Find the possible values of \(p\) fithe explession \(4 x^{2}-3 p x+3 p\) leaves a reminder 10 when divided by \(x-p\)
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Correct Answer: Option A
Explanation:
Let \(f( x )=4 x ^{2}-3 px +3 p\)
Then, when \(f( x )\) is divided by \(x - p\), the remainder is \(f( p )\)
\begin{aligned}
&\therefore f(p)=4 p^{2}-3 p \times p+3 p=10 \\
&4 p^{3}-3 p^{2}+3 p=10 \\
&P^{2}+3 p-10=0 \\
&P^{2}+5 p-2 p-10=0 \\
& P ( p +5)-2(p+5)=0 \\
&(p+5)(p-2)=0 \\
&\Rightarrow p+5=0 \text { or } p-2=0 \\
&\Rightarrow p =15 \text { or } 2
\end{aligned}
Let \(f( x )=4 x ^{2}-3 px +3 p\)
Then, when \(f( x )\) is divided by \(x - p\), the remainder is \(f( p )\)
\begin{aligned}
&\therefore f(p)=4 p^{2}-3 p \times p+3 p=10 \\
&4 p^{3}-3 p^{2}+3 p=10 \\
&P^{2}+3 p-10=0 \\
&P^{2}+5 p-2 p-10=0 \\
& P ( p +5)-2(p+5)=0 \\
&(p+5)(p-2)=0 \\
&\Rightarrow p+5=0 \text { or } p-2=0 \\
&\Rightarrow p =15 \text { or } 2
\end{aligned}