: If the expression Px² + 3x2 - 3 and 2x? - 5x+P when divided
by x – 4 leave the same remainder, find the value of P.
Answers
Answer:
Given, f(x) = ax³+3x²-3
\: \: \: \: \: \: g(x) = 2 {x}^{3} - 5x + ag(x)=2x
3
−5x+a
\begin{gathered} If \: f(x) \: is \: divided \: by \: (x-4) \\ then \: it \:leaves \: a \: remainder \: f(4) \end{gathered}
Iff(x)isdividedby(x−4)
thenitleavesaremainderf(4)
\begin{gathered}f(4) = a \times 4 ^{3} + 3(4 ^{2} ) - 3 \\ = a \times 64 + 48 - 3 \\ = 64a + 45\end{gathered}
f(4)=a×4
3
+3(4
2
)−3
=a×64+48−3
=64a+45
\begin{gathered} If \: g(x) \: is \: divided \: by \: (x-4) \\ then \: it \:leaves \: a \: remainder \: g(4) \end{gathered}
Ifg(x)isdividedby(x−4)
thenitleavesaremainderg(4)
\begin{gathered}g(4) = 2(4 ^{3} ) - 5 \times 4 + a \\ = 128 - 20 + a \\ = 108 + a \\ \\ \\ Given \: that \: remainders \: in \: both \: \\ cases \: are \: equal \: \\ \\ Now, \: \\ 108 + a = 64a + 45 \\ 63a = 108 - 45 = 63 \\ \\ a = \frac{63}{63} = 1\end{gathered}
g(4)=2(4
3
)−5×4+a
=128−20+a
=108+a
Giventhatremaindersinboth
casesareequal
Now,
108+a=64a+45
63a=108−45=63
a=
63
63
=1