Question

If P(x) is a polynomial of degree 4 with leading coefficient as three such that P(1)=2, P(2)=8, P(3)=18, P(4)=32, then the value of P(5) is :
A.120
B.121
C.122
D.123

Answer 1

Hint: We are given a polynomial of degree 4 that is the highest power of x in the polynomial P(x) is 4. The leading coefficient is the coefficient of the term of the highest degree. With the help of this information, form the polynomial equation and then find the coefficient of the rest of the terms to find the correct answer.

Complete step-by-step answer:
A polynomial function is defined as a single or a sum of algebraic terms having different powers of the same variable.
The degree of a polynomial is the highest power present in it.
For example, $ 5{x^4} + 2{x^2} - 8 $ is a polynomial equation whose degree is 4, the coefficient of $ {x^4} $ is 5 and that of $ {x^2} $ is 2.
The general form of a polynomial equation of degree 4 is $ A{x^4} + B{x^3} + C{x^2} + Dx + E = P(x) $ .
Now we are given that the leading coefficient is 3, then the polynomial equation becomes,
 $ P(x) = 3{x^4} + B{x^3} + C{x^2} + Dx + E $
We are given that P(1) = 2 that is,
 $
  3{(1)^4} + B{(1)^3} + C{(1)^2} + D(1) + E = 2 \\
  3 + B + C + D + E = 2 \\
  B + C + D + E = - 1...(1) \;
  $
P(2) = 8 that is,
 $
  3(16) + B(8) + C(4) + D(2) + E = 8 \\
  48 + 8B + 4C + 2D + E = 8 \\
  8B + 4C + 2D + E = - 40...(2) \;
  $
P(3) = 18 that is,
 $
  3(81) + B(27) + C(9) + D(3) + E = 18 \\
  243 + 27B + 9C + 3D + E = 18 \\
  27B + 9C + 3D + E = - 225...(3) \;
  $
P(4) = 32 that is,
 $
  3(256) + B(64) + C(16) + D(4) + E = 32 \\
  768 + 64B + 16C + 4D + E = 32 \\
  64B + 16C + 4D + E = - 736...(4) \;
  $
There are 4 variables and 4 equations to find their value, we solve the equations by elimination method –
 $
  (2) - (1) \\
  8B + 4C + 2D + E = - 40 \\
  \underline { - B - C - D - E = + 1} \\
  \underline {7B + 3C + D = - 39\,\,\,\,} ...(5) \;
  $
 $
  (3) - (1) \\
  64B + 16C + 4D + E = - 736 \\
  \underline { - B - C - D - E = + 1\,\,\,\,\,} \\
  \underline {63B + 15C + 3D = - 735} \\
   \Rightarrow 21B + 5C + D = - 245...(6) \;
  $
 $
  (4) - (1) \\
  27B + 9C + 3D + E = - 225 \\
  \underline { - B - C - D - E = + 1\,\,\,} \\
  \underline {26B + 8C + 2D = - 224} \\
   \Rightarrow 13B + 4C + D = - 112...(7) \;
  $
Now,
 $
  3 \times (5) - (6) \\
  21B + 9C + 3D = - 117 \\
  \underline { - 21B - 5C - D = + 245} \\
  \underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4C + 2D = 128} \\
   \Rightarrow 2C + D = 64...(8) \;
  $
And
 $
  7 \times (7) - 13(5) \\
  91B + 28C + 7D = - 784 \\
  \underline { - 91B - 39C - 13D = + 507} \\
  \underline { - 11C - 6D = - 277} \\
   \Rightarrow 11C + 6D = 277...(9) \\
  $
 $
  (9) - 6 \times (8) \\
  11C + 6D = 277 \\
  \underline { - 12C - 6D = - 384} \\
  \underline {\,\,\,\,\,\,\,\,\,\,\,\, - C = - 107\,\,\,\,} \\
   \Rightarrow C = 107 \;
  $
Put the value of C in (8)
 $
  2(107) + D = 64 \\
  D = - 150 \;
  $
Put the value of C and D in (5)
 $
  7B + 3(107) + ( - 150) = - 39 \\
  7B = - 210 \\
   \Rightarrow B = - 30 \;
  $
Put the value of B, C and D in (1)
 $
   - 30 + 107 - 150 + E = - 1 \\
   \Rightarrow E = 72 \;
  $
Thus, we get,
 $ B = - 30 $ , $ C = 107 $ , $ D = - 150 $ and $ E = 72 $
Hence the polynomial is,
 $ P(x) = 3{x^4} - 30{x^3} + 107{x^2} - 150x + 72 $
Now we have to find the value of P(5) ,
 $
  P(5) = 3(625) - 30(125) + 107(25) - 150(5) + 72 \\
  P(5) = 122 \;
  $
So, the correct answer is “Option C”.

Note: An expression containing square roots of variables, fraction or negative powers on the variables and variables are present in the denominator of any fraction, the expression cannot be classified as a polynomial. Also, a function can have as many zeros as the degree of the polynomial.