Question

Let f(x) =
${x}^{5} + a {x}^{4} + b {x}^{3} + c {x}^{2} + d$
such that f(1) = 1, f(2) = 2, f(3) = 3 ,f(4)= 4 ,f(5)= 5, Then,what is the value of d.

Answer 1

Hi friend, Harish here.

Here is your answer:

Given that:

$f(x)= {x}^{5} + a {x}^{4} + b {x}^{3} + c {x}^{2} + d$

$f(1)=1, f(2)=2, f(3)=3,f(4)=4,f(5)=5.$

To find:

The value of d.

Solution:

f(x) can be written as :

$f(x)=(x-1)(x-2)(x-3)(x-4)(x-5) + x$  - (i)

⇒ $f(x) = {x}^{5} + a {x}^{4} + b {x}^{3} + c {x}^{2} +d$  - (ii)

$In\ (ii)\ equation, \ f(0) = d .$

Now substituting x value in (i) equation we get;

$f(0) = (-1)(-2)(-3)(-4)(-5) + 0 = -120.$

Therefore the value of d = -120.
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Hope my answer is helpful to you. 

Answer 2

f(x) = x⁵+ax⁴+bx³+cx²+d
f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4, f(5) = 5
f(x) can be written as f(x) = (x-1)(x-2)(x-3)(x-4)(x-5) + x
Hence, f(0) = d = -120