Question

Interesting Question!!

A quartic polynomial P(x) satisfies the following properties.
(A) P(3)=32
(B) P(2x+1)=16P(x+1) for all real x .
Then, find P(4).


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Answer 1

Step-by-step explanation:

p(3)= 32

p(2x+1) = 16 p( x+1)

p(4)= p(3) +p(1)

= 32 + p(2×0+1) = 32 + 16 ×p(0+1)

= 32 + 16× p(1)

here, p(1) =?

p(3) = p( 2×1 +1) = 16×p(2)=..

Answer 2

$\large\underline{\text{Solution}}$

We know that we can find the value of $P(1)$ by direct substitution of $x=0$.

$\hookrightarrow P(1)=16P(1)$

$\large\hookrightarrow\red{\boxed{\red{\bold{P(1)=0}}}}$

By factor theorem, $(x-1)$ is a factor of $P(x)$.

So let us assume,

$\hookrightarrow P(x)=(x-1)^{n}Q(x),\ Q(1)\neq0$

By assumption,

$\large\hookrightarrow \red{\boxed{\red{\bold{P(2x+1)=(2x)^{n}Q(2x+1)}}}}$

$\large\hookrightarrow \red{\boxed{\red{\bold{16P(x+1)=16x^{n}Q(x+1)}}}}$

For any real $x$,

$\hookrightarrow(2x)^{n}Q(2x+1)=16x^{n}Q(x+1),\ Q(1)\neq0$

As $x\neq0$ satisfies the equation,

$\hookrightarrow2^{n}Q(2x+1)=16Q(x+1),\ Q(1)\neq0$

By substituting $x=0$,

$\hookrightarrow2^{n}Q(1)=16Q(1),\ Q(1)\neq0$

$\hookrightarrow 2^{n}=16$

$\large\hookrightarrow\red{\boxed{\red{\bold{n=4}}}}$

Since $P(x)$ and $(x-1)^{n}$ are both a quartic polynomial, $Q(x)$ is a constant term.

Let us assume $Q(x)=k$.

By assumption,

$\hookrightarrow P(3)=32$

$\hookrightarrow 16k=32$

$\large\hookrightarrow\red{\boxed{\red{\bold{k=2}}}}$

Hence,

$\large\hookrightarrow\red{\boxed{\red{\bold{P(x)=2(x-1)^{4}}}}}$

And hence,

$\large\hookrightarrow\red{\boxed{\red{\bold{P(4)=2\times3^{4}=\underline{162}}}}}$

This is our answer.