Garvil's polynomial of degree 4

Construct a new monic polynomial $p(x)$ that satisfies $p(x)=f(x)-10x$ .

It is obvious that $p(x)$ is a four degree polynomial.

Substitute $x$ with $1,2,$ and $3$ , we get $p(x)=0$ . Thus $1,2$ and $3$ the roots of $p(x)$ .

Let $a$ be the fourth root of $p(x)$ , then $p(x)=(x-a)(x-1)(x-2)(x-3)$ .

$p(x)=f(x)-10x$

$\iff f(x)=p(x)+10x$

$\iff f(x)=(x-a)(x-1)(x-2)(x-3)+10x$

Then, $f(12)+f(-8)-19000$

$=[(12-a)(12-1)(12-2)(12-3)+10 \cdot 12]+[(-8-a)(-8-1)(-8-2)(-8-3)+10 \cdot (-8)]-19000$

$=[11 \cdot 10 \cdot 9 \cdot (12-a)+120]+[(-8-a)(-9)(-10)(-11)-80]-19000$

$=[990(12-a)+120]+[-990(-8-a)-80]-19000$

$=990(12-a-(-8-a))+(120-80)-19000$

$=990 \cdot 20+40-19000$

$=840$

Let $f(x)=x^4+ax^3+bx^2+cx+d$

Then:

$f(1)=a+b+c+d+1 =10 \\ f(2)=8a+4b+2c+d+16 =20 \\ f(3)=27a+9b+3c+d+81=30$

To answer the question, we would like to find values $r_1, r_2, r_3, r_4$ such that:

$f(12)+f(-8)-19000 = r_1f(1) + r_2f(2) + r_3f(3) + r_4$

Substituting for $f(x) = x^4 + ax^3 + bx^2 + cx+d$ , we obtain

$1216a + 208b + 4c + 2d + 5832 = r_1f(1) + r_2f(2) + r_3f(3) + r_4 \\ = r_1 (a+b+c+d+1) \\ + r_2 (8a+4b+2c+d+16) \\ + r_3(27a+9b+3c+d+81) \\ + r_4 \\$

Comparing the coefficients of $a, b, c, d$ , this leads to the following system of equations:

$r_1+8r_2+27r_3=1216 \\ r_1+4r_2+9r_3=208 \\ r_1+2r_2+3r_3=4 \\ r_1+r_2+r_3=2 \\ r_4=5832-(r_1+16r_2+81r_3)$

Solving this system, we get: $(r_1,r_2,r_3,r_4) = (100,-198,100,800)$

Plugging these results into the expression before:

$f(12)+f(-8)-19000=100f(1)-198f(2)+100f(3)+800= \boxed {840}$

Let g(x) = f(x) - 10x. Then g(1) = g(2) = g(3) = 0, so g(x) = (x-1)(x-2)(x-3) p(x) for some p(x). Since g(x) is also monic and quartic, p(x) = x-c for some c. Then g(12) + g(-8) = p(12) (11)(10)(9) + p(-8) (-9)(-10)(-11) = (p(12)-p(-8) 990 = 20*990 = 19800, so f(12) + f(-8) = g(12) + g(-8) + 120 - 80 = 19840, for a final answer of 840.

Let g defined by $g(x) = f(2 - x) + f(2 +x)$

We have $g(x) = g(-x)$ for every $x$

Then coefficients for odd power of $x$ are null.

So $g(x)$ can be writen $g(x) = 2x^4 + ax^2 + b$

We have $g(0) = 40 \implies b = 40$

and $g(1) = 40 \implies 2 + a + b = 40 \implies a = - 2$

and finally $f(12) + f(-8) -19000 = g(10) - 19000 = 840$

Let $f(x) = x^4+ax^3+bx^2+cx+d$ , then we have:

$\begin{cases} f(1) = 1 + a + b + c + d = 10 & \Rightarrow a + b + c + d = 9 &...(1) \\ f(2) = 16 + 8a + 4b + 2c + d = 20 & \Rightarrow \color{#3D99F6}{8a + 4b + 2c + d = 4} &...(2) \\ f(3) = 81 + 27a + 9b + 3c + d = 30 & \Rightarrow 27a + 9b + 3c + d = -51 &...(3) \end{cases}$

$\begin{cases} (2)-(1): & 7a + 3b + c = -5 & ...(4) \\ (3)-(2): & 19a + 5b + c = -55 &...(5) \\ (5)-(4): & \color{#D61F06}{12a + 2b = -50} &...(6) \end{cases}$

$\begin{aligned} f(12) + f(-8) - 19000 & = 20736 + 1728a + 144b + 12c + d \\ & \quad +4096 - 512a + 64b - 8c + d - 19000 \\ & = 24832 + 1216a + 208b + 4c + 2d - 19000 \\ & = 5832 +\color{#D61F06}{1200a}+\color{#3D99F6}{16a} + \color{#D61F06}{200b}+\color{#3D99F6}{8b} + \color{#3D99F6}{4c}+\color{#3D99F6}{2d} \\ & = 5832 +\color{#D61F06}{100(-50)}+\color{#3D99F6}{2(4)} \\ & = \boxed{840} \end{aligned}$

In a monic polynomial, the leading coefficient is 1 So let f(x) = x^4 + ax^3 + bx^2 + cx + d

f(1) = 1 +a+b+c+d = 10

a+b+c+d = 9 , (i)

f(2) = 16 + 8a + 4b + 2c + d = 20

8a + 4b + 2c + d = 4 , (ii)

f(3) = 81 + 27a + 9b + 3c + d = 30

27a + 9b + 3c + d = -51 , (iii)

Now, (ii)-(i)

7a + 3b + c = -5, (iv)

Now, (iii)-(i):

26a + 8b + 2c = -60 13a + 4b + c = -30 , (v)

Now, (v)-(iv):

6a + b = -25

we cannot find explicit values of a, b, c, and d, since we have only three equations with four unknowns.

f(12) + f(-8) - 19000

= 12^4 + 12^3 a + 12^2 b + 12c + d + (-8)^4 + (-8)^3 a + (-8)^2 b - 8c + d - 19000

= 20736 + 1728a + 144b + 12c + d + 4096 - 512a + 64b - 8c + d - 19000

= 5832 + 1216a + 208b + 4c + 2d

= 5232 + 2(608a + 104b + 2c + d)

I was hoping for one of the above relations to show up , but no such luck. we could try to back-substitute, e.g. b = -25-6a but that back into (iv) and get "c" in terms of "a" etc. If we solve a,b,c in terms of d, we get, a=-(d+36)/6,

b=d+11,

c=-(11d-24)/6

from which we can define

f(x)=x^4 -(d+36)x^3/6 +(d+11)x^2 -(11d-24)x/6 +d

and based on this,

f(1)=10

Hence, Putting the value of x as 1 in the above f(x) we get d=-48

f(12)+f(-8)-19000

=-608(d+36)/3 +208(d+11)+2d+2(24-11*d)/3 +5832

Now putting the value of d=-48 we get :-

=840 <<==ANSWER

Known : Let, f(x) = x^4 + ax^3 + bx^2 + cx + d from f(1), a+b+c+d = 9.................p1 f(2), 8a+4b+2c+d = 4...................p2 f(3), 27a+9b+3c+d = -51.............p3

p2-p1 = 7a+3b+c = -5..................p4 p3-p2 = 19a+5b+c = -55..............p5

p5-p4 = 12a+2b=-50 => 6a+b=-25...................p6

Question : f(12) + f(8) -19000...? Ans : f(12) + f(-8) -19000 = (12^4 + 12^3a + 12^2b +12c + d) + ((-8)^4 + (-8)^3a + (-8)^2b + (-8)c + d) - 19000 = (12^3 - 8^3)a + (12^2 + 8^2)b + (12-8)c + 2d + 20736 + 4096 - 19000 = 1216a + 208 b + 4c + 2d + 5832...............p7 from p2 => 2c + d = 4 - 8a - 4b so 4c + 2d = 8 -16a - 8b Subs to p7 =1216a + 208b + 8 -16a - 8b + 5832 =1200a + 200b + 5840 =200(6a+b) + 5832, from p6 = 6a+b=-25 =200(-25) + 5840 =-5000 + 5840 =840

Assume f(x) = x^4 + a x^3 + b x^2 + c*x + d

substituting 1,2,3 we get

a + b + c + d = 9 8a + 4b + 2c +d = 4 27a + 9b + 3c +d = -51

we need one more equation to get a,b,c,d

since from given data f(3) = f(2) + f(1)

so we obtain 18a + 4b + d = -64

solving them by matrix method we obtain

a = -6, b = 11, c = 4, d = 0 thus we get the solution of final equation.

This is a very long method if any shortcut is obtained please let me know. And sorry if any data values go wrong.

At first glance, it looks as if the problem is incomplete, as "a monic polynomial $f(x)$ of degree four" gives us four degrees of freedom, but only three of these are absorbed by the values given for specific values of $x$ . As such, we have to try and reduce the equations, and hope that the extra degree of freedom vanishes in the formula that we are given to calculate.

Let $f(x) = x^4 + ax^3 + bx^2 + cx + d \\ Z = f(12) + f(-8) - 19000$ Then $1 + a + b + c + d = 10 \Rightarrow a + b + c + d = 9 \\ 16 + 8a + 4b + 2c + d = 20 \Rightarrow 8a + 4b + 2c + d = 4 \\ 81 + 27a + 9b + 3c + d = 30 \Rightarrow 27a + 9b + 3c + d = -51\\ Z = 20736 + 1728a + 144b + 12c + d + 4096 - 512a + 64b - 8c + d - 19000 \\ = 1216a + 208b + 4c + 2d + 5832$ Now we reduce this system of equations by substitution. We now know that $d = 9 - a - b - c$ , which gives us $9 = 9 7a + 3b + c + 9 = 4 \Rightarrow 7a + 3b + c = -5 \\ 26a + 8b + 2c + 9 = -51 \Rightarrow 13a + 4b + c = -30 \\ Z = 1216a + 208b + 4c + (18 - 2a - 2b - 2c) + 5832 \\ = 1214a + 206b + 2c + 5850$ This tells us that now know that $c = -5 - 7a - 3b$ , giving us $-5 = -5 \\ 6a + b - 5 = -30 \Rightarrow 6a + b = -25 \\ Z = 1214a + 206b + (-10 - 14a - 6b) + 5850 \\ = 1200a + 200b + 5840$ so (b = -25 - 6a), giving $Z = 1200a + 200b + 5840 \\ = 1200a - 5000 - 1200a + 5840 \\ = \boxed{840}$

We can use Newton's forward difference interpolation formula which states that $f(x) = f(x_0) + \frac{(x-x_0)}{h}\Delta f(x_0) + \frac{(x-x_0)(x-x_1)}{2!h^2}\Delta^2 f(x_0) + \frac{(x-x_0)(x-x_1)(x-x_2)}{3!h^3}\Delta^3f(x_0) + \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)}{4!h^4}\Delta^4f(x_0)$

for a fourth degree polynomial where h is the difference between the x values given. I will take $x_0 = 1$ so using a forward differences table $f(x_0) = 10$

$\Delta f(x_0) = 10$

$\Delta^2 f(x_0) = 0$ .

We also know that the required polynomial is monic so the coefficient of the $x^4$ term is 1. In the formula, the only $x^4$ comes from the last term, of which the numerator of the fraction contains a $x^4$ . This means that $\frac{\Delta^4 f(x_0)}{4!h^4} = 1$ . h is 1 because we know the values of $f(x)$ at x values of 1, 2, 3 which is an arithmetic progression of difference 1. Therefore it follows that $\Delta^4 f(x_0) = 24$ .

Let the value of $\Delta^3 f(x_0)$ be m.

The polynomial is therefore $f(x) = 10 + 10(x-1) + \frac{m}{6}(x-1)(x-2)(x-3) + (x-1)(x-2)(x-3)(x-4)$

This rearranges to $f(x) = x^4 + (\frac{m}{6} - 10)x^3 + (35-m)x^2 (\frac{11m}{6} - 40)x + (24-m)$

Using this, $f(12) = 20736+288m-17280+5040-144m+22m-480+24-m = 8040 + 165m$

$f(-8) = 4096 - \frac{256}{3}m + 5120+2240-64m-\frac{44}{3}m +320 + 24-m = 11800 - 165m$

When these two are added together, the $m$ s cancel out and we get 19840. Subtracting 19000 gives the answer of $\boxed{840}$ .

As we have $f(1)$ , $f(2)$ and $f(3)$ , we could use polynomial interpolation and obtain a polynomial

$g(x) = 10 {x-1 \choose 0} + 10 {x-1 \choose 1} + 0 {x-1 \choose 2}$ .

If we assume: $f(x) = g(x) + \alpha {x-1 \choose 4}$ in order to obtain a monic polynomial of degree four, is necessary to determine $\alpha$ such that the coefficient of the term $x ^ 4$ is $1$ . It's clear to see that $\alpha = 4! = 24$ .

Thus,

$f(x) = 10 {x-1 \choose 0} + 10 {x-1 \choose 1} + 0 {x-1 \choose 2} + 24 {x-1 \choose 4}$ .

So, $f(12) + f(-8) - 19000 = 8040 + 11800 - 19000 = 840$ .

Let a general fourth degree monic polynomial with coefficients 1,a,b,c and d for x^4, x^3, x^2, x and x^0 respectively. Then insert given data in the polynomial and solve the system of linear equations. I found a=-6, b=11, c=3 and d=0. Thus I found the fourth degree polynomial and calculated f(12)+f(-8)-1900

Since we know the leading coefficient is 1 and the degree is 4. We can then suppose the function is $f(x)=x^4+ax^3+bx^2+cx+d$ , where $a,b,c,d$ are undetermined coefficients. Since $f(1)=10, f(2)=20, f(3)=30$ , we can then solve the miscellaneous equations in terms of $a$ . We get $b=-6a-25, c=11a+70, d=-6a-36$ . Because of the generality of the question, we can simply substitute a value for $a$ and then work out the values of $b,c,d$ . Here, I let $a$ become 1, and $b=-31,c=81,d=-42$ And our function becomes $f(x)=x^4+x^3-31x^2+81x-42$ And let $x=12$ and $x=-8$ , we get $f(12)=18930, f(-8)=910$ Therefore $f(12)+f(-8)-19000=840$ .

a=-5 b=5 c=15 d=-6 and hence put the values to get the desired answer