Let $P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$ where $a,b,c,d \in R$.
Suppose $P\left( 0 \right) = 6,P\left( 1 \right) = - 7,P\left( 2 \right) = 8$ and $P\left( 3 \right) = - 9$, then find the value of $P\left( 4 \right)$.

Hint:
Here, we will find the values of the coefficients by substituting each given value in the equation one by one. Then we will get a set of linear equations containing only coefficients. We will solve these linear equations to find all the values. Then will reframe the given polynomial by substituting the values of the coefficients. We will then substitute 4 in the reframed polynomial in place of variable to get the required value.

Complete step by step solution:
The polynomial P is given as
$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$…………………..$\left( 1 \right)$
We have four points of x at which the polynomial P value is given so we will form four equations as follows.
Firstly, At $x = 0$ $P\left( x \right) = 6$
So substituting above value in equation $\left( 1 \right)$ we get,
$6 = 0 + a \times 0 + b \times 0 + c \times 0 + d$
Simplifying the expression, we get
$ \Rightarrow d = 6$
Next, At $x = 1$$P\left( x \right) = - 7$
So substituting above value in equation $\left( 1 \right)$ we get
$ - 7 = {1^4} + a \times {1^3} + b \times {1^2} + c \times 1 + d$
$ \Rightarrow - 7 = 1 + a + b + c + d$
Substituting $d = 6$ in above equation, we get
$\begin{gathered}
   \Rightarrow - 7 = 1 + a + b + c + 6 \\
   \Rightarrow a + b + c = - 7 - 1 - 6 \\
\end{gathered} $
Adding the like terms, we get
$ \Rightarrow a + b + c = - 14$……………………………$\left( 3 \right)$
Next, at $x = 2$$P\left( x \right) = 8$
Substituting above value in equation $\left( 1 \right)$ we get
$ 8 = {2^4} + a \times {2^3} + b \times {2^2} + c \times 2 + d \\
   \Rightarrow 8 - {2^4} = 8a + 4b + 2c + d \\ $
Substituting $d = 6$ in above equation and solving, we get
$ \Rightarrow 8 - 16 = 8a + 4b + 2c + 6$
$ \Rightarrow 8a + 4b + 2c = - 14$
The above equation can further simplify by dividing it by 2.
$4a + 2b + c = - 7$……………………….$\left( 4 \right)$
Next, at $x = 3$$P\left( x \right) = - 9$
So, substituting above value in equation $\left( 1 \right)$we get
$\Rightarrow - 9 = {3^4} + a \times {3^3} + b \times {3^2} + c \times 3 + d \\
   \Rightarrow - 9 - {3^4} = 27a + 9b + 3c + d \\ $
Substituting $d = 6$ in above equation and solving, we get
$ \Rightarrow - 9 - 81 = 27a + 9b + 3c + 6 \\
   \Rightarrow 27a + 9b + 3c = - 96 \\ $
The above equation can further simplify by dividing it by 3.
$9a + 3b + c = - 32$…………………………$\left( 5 \right)$
Now our next step will be solving the equation $\left( 3 \right)$, $\left( 4 \right)$ and $\left( 5 \right)$ to get the value of $a,b,c$.
Subtracting equation $\left( 3 \right)$ from $\left( 5 \right)$ we get
$\left( {9a + 3b + c} \right) - \left( {a + b + c} \right) = - 32 - \left( { - 14} \right) \\
   \Rightarrow 9a - a + 3b - b + c - c = - 32 + 14 \\ $
On solving, we get
$8a + 2b = - 18$………………………………$\left( 6 \right)$
Subtracting equation $\left( 5 \right)$ from $\left( 4 \right)$ we get
$\left( {9a + 3b + c} \right) - \left( {4a + 2b + c} \right) = - 32 - \left( { - 7} \right) \\
   \Rightarrow 9a - 4a + 3b - 2b + c - c = - 32 + 7 \\ $
On solving we get
$5a + b = - 25$………………………………$\left( 7 \right)$
Multiplying equation $\left( 7 \right)$ by 2 and subtracting it by equation $\left( 6 \right)$ we get
$2\left( {5a + b} \right) - \left( {8a + 2b} \right) = 2 \times - 25 - \left( { - 18} \right) \\
   \Rightarrow 10a + 2b - 8a - 2b = - 50 + 18 \\$
On solving, we get
$ 2a = - 32 \\
   \Rightarrow a = - 16 \\ $
Putting value of $a$ in equation $\left( 6 \right)$, we get
$ 8 \times - 16 + 2b = - 18 \\
   \Rightarrow 2b = - 18 + 128 \\
   \Rightarrow b = 55 \\ $
Substituting value of $a$ and $b$ in equation $\left( 3 \right)$ we get
$- 16 + 55 + c = - 14 \\
   \Rightarrow c = - 14 + 16 - 55 \\
   \Rightarrow c = - 53 \\ $
So we get our values as
$ a = - 16 \\
  b = 55 \\
  c = - 53 \\
  d = 6 \\ $
Substituting above value in equation $\left( 1 \right)$ we get
$P\left( x \right) = {x^4} + {x^3} \times - 16 + {x^2} \times 55 + x \times \left( { - 53} \right) + 6$
$ \Rightarrow P\left( x \right) = {x^4} - 16{x^3} + 55{x^2} - 53x + 6$…………………………..$\left( 8 \right)$
Now, we have to find the value of the polynomial at $x = 4$.
Putting $x = 4$ in equation (8), we get
$P\left( x \right) = {4^4} - 16 \times {4^3} + 55 \times {4^2} - 53 \times 4 + 6$
Applying the exponent on the terms, we get
  $ \Rightarrow P\left( x \right) = 256 - 1024 + 880 - 212 + 6 \\
   \Rightarrow P\left( x \right) = - 94 \\ $

Note:
We should always check whether the number of equations is equal to the number of variables or not. We can find the values of all the variables only if the number of variables is equal to the number of equations. We have framed a linear equation using coefficient only. A linear equation is an equation which has a highest degree of 1.