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# Help with polynomial

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Let f(x) = x^4+ax^3+bx^2+cx+d, where a,b,c,and d are real numbers. Suppose the graph of y=f(x) intersects the graph of y=2x+3 at x =1,2,3. What is the value of f(0) + f(4)?

Jun 14, 2021

#1
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It seems there's not enough information to find \$f(0)+f(4)\$;  I need to know the value of \$f(x)\$ at one more point.

Jun 15, 2021
#2
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We must have \({x}^{4}+a{x}^{3}+b{x}^{2}+cx+d=2x+3\) at x =1, x=2, and x=3. Substituting we get

a + b + c +d = 4                                             (1)

8a + 4b + 2c + d = - 9                                     (2)

27a + 9b + 3c +d = - 72                                   (3)

Subtracting (1) from (2) and (2) from (3) gives

7a +3b + c = -13                                              (4)

19a +5b +c = - 63                                            (5)

Subtracting (4) from (5) results in

12a + 2b = - 50                                               (6)

f(0) + f(4)= d + 256 + 64a + 16b + 4c + d

= 256 + 64a + 16b + 4c + 2d                         (7)

Solving (2) for 2c + d we get

2c + d = - 9 - 8a - -4b                                   (8)

Substituting (8) in 7 gives

f(0) + f(4) = 256 +64a +16b +2(- 9 - 8a - 4b)

= 256 +64a +16b - 18 -16a - 8b

= 238 +48a + 8b

= 238 +4(12a +2b)                                        (9)

Substituting for 12a +2b the value we found in (6) completes the lengthy computations:

f(0) + f(4) = 238 + 4(-50) = 38.

Unless I am in error, no ther value of the function is needed.

Jun 15, 2021