Math, asked by Anonymous, 1 month ago

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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Answers

Answered by jaseelak519
0

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)=..

Answered by user0888
5

\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.

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