Question

f(x) is a biquadratic polynomial whose leading coefficient is 1 and f(1)=1, f(2)=4, f(3)=9, f(4)=16. Find the biquadratic polynomial​

Answer 1

Answer:

.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Let f(x)=ax³+bx²+cx+d

f(1)=a+b+c+d=1------------------(1)

f(2)=8a+4b+2c+d=2-------------(2)

f(3)=27a+9b+3c+d=3-----------(3)

f(4)=64a+16b+4c+d=16--------(4)

subtracting (1) from (2),(3) and (4) respectively we have,

7a+3b+c=1-------------------(5)

26a+8b+2c=2----------------(6)

63a+15b+3c=15-------------(7)

multiplying (5) with 2 and subtracting from (6) and multiplying (5) with 3 and subtracting from (7) we have,

12a+2b=0-------(8)

42a+6b=12-----(9)

multiplying (8) with 3 and (9) with 1 and subtracting we have,

36a+6b=0

42a+6b=12

6a=12

or, a =12

then, from (8) we have, 2b=-12a=-12×12=-144

or, b=-144/2=-72

then from (5), c=1-3b-7a=1-3×(-72)-7×12=1+216-84=217-84=133

then from (1), d=1-a-b-c=1-12-(-72)-133=1-12+72-133=73-145=-72

then, f(x)=12x³-72x²+133x-72

∴, f(5)=12×5³-72×5²+133×5-72

=12×125-72×25+665-72

=1500-1800+593

=2093-1800

=293