P(x)=4x^3-12x^2+14x-3,g(x)=2x-1-1. By remainder therom
Answers
PLEASE MARK AS BRAINLIST
FIRSTLY we will take gx and put it equal to 0
2x-1-1=0
Then x=1
Now put the value of x in the px
Solve it as the value of x is 1... Whatever comes at last is the remainder
Answer:
Answer:
By remainder theorem, we know that p(x) when divides by
\begin{gathered} \boxed{ \sf g(x) = \bigg(x - \frac{1}{2} \bigg)} \\ \end{gathered}
g(x)=(x−
2
1
)
gives a remainder equal to
\begin{gathered} \boxed {\sf p \bigg( \frac{1}{2} \bigg)} \\ \\ \end{gathered}
p(
2
1
)
Now,
p(x) = 4x³-12x²+14x-3
\begin{gathered} \\ \implies \sf p \bigg( \frac{1}{2} \bigg) = 4 \bigg( \frac{1}{2} \bigg) {}^{3} - 12 \bigg( \frac{1}{2} \bigg) {}^{2} \\ \\ \sf + 14 \bigg( \frac{1}{2} \bigg) - 3 \\ \\ \\ \implies \sf \frac{4}{8} - \frac{12}{4} + \frac{14}{2} - 3 \\ \\ \\ \implies \sf \frac{1}{2} - 3 + 7 - 3 \\ \\ \\ \implies \sf \blue{ \frac{3}{2} } \\ \\ \end{gathered}
⟹p(
2
1
)=4(
2
1
)
3
−12(
2
1
)
2
+14(
2
1
)−3
⟹
8
4
−
4
12
+
2
14
−3
⟹
2
1
−3+7−3
⟹
2
3
Hence, the required remainder :
\begin{gathered} \boxed{ \sf p \bigg( \frac{1}{2} \bigg) = \frac{3}{2} } \\ \\ \end{gathered}
p(
2
1
)=
2
3