Evaluate at 5

Let $f$ be a polynomial of degree $n$ . $f(x)=a_n x^n+a_{n-1} x^{n-1}+\ldots+a_1 x+a_0$ . So $f(x+2)=a_n(x+2)^n+a_{n-1} (x+2)^{n-1}+\ldots+a_{1}(x+2)+a_0$ $=a_n x^n+a_n \cdot {n \choose 1}(2x^{n-1})+a_n \cdot {n \choose 2}(2^2x^{n-2})+\ldots$ $+a_{n-1} x^{n-1}+a_{n-1} \cdot {n \choose 1}(2x^{n-2})+a_{n-1} \cdot {n \choose 2}(2^2x^{n-3})+\ldots$ . Since $ax^n$ is the only term with exponent $n$ in both $f(x)$ and $f(x+2)$ , so in $f(x+2)-f(x)$ , the term $ax^n$ will cancel out, and the degree of $f(x+2)-f(x)$ is $n-1=2$ . Hence, $n=3$ .

Let $f(x)=ax^3+bx^2+cx+d$ . Then $f(0)=d=-16$ . Also, $f(x+2)-f(x)=6ax^2+12ax+8a+4bx+4b+2c=36x^2+48x+16$ . By comparing coefficients, $6a=36,12a+4b=48,8a+4b+2c=16$ , which gives $a=6,b=-6,c=-4$ . So $f(x)=6x^3-6x^2-4x-16$ , and $f(5)=564$ .

[Notation edits. He let $f(x)=ax^n+bx^{n-1}+\ldots+mx+n$ , what is wrong with that? - Calvin]

since f(x+2) - f(x) = a quadratic expression

so f(x) has to be a cubic polynomial. so f(x) =a x^3 + b x^2 + c*x +d

since f(o)= -16 then d = -16

f(x+2) -f(x) = 2a(3 x^2 +6x+4)+ b(4x+4) + 2c = 6a x^2 + (12a + 4b)*x + 2c+4a+4b Now equating coefficients with (6x+4)^2

we have a = 6; b= -6; c = -4

so f(x) = 6 x^3 - 6 x^2 -4x -16

now f(5) = 564

If the polynomial $f(x)$ had a degree higher than $3$ , then the difference $f(x+2)-f(x)$ would have higher powers of $x$ than $2$ . So let's assume $f(x) = ax^3+bx^2+cx+d$ .

As $f(0) = d$ which is $-16$ , so $f(x) = ax^3+bx^2+cx -16$ .

Now, from $f(x+2)-f(x) = (6x+4)^2$ , we get, $a(x+2)^3+b(x+2)^2+c(x+2)-16 -$ $(ax^3+bx^2+cx-16) = 36x^2+48x+16$ , or simplifying, $6 a x^2+12 a x+8 a+4 b x+4 b+2 c=36 x^2+48 x+16$ , from where we can get this three equations by equating the coefficients of $x^2, x^1 , x^0$ on both sides,

$6a = 36$

$12a+4b = 48$

$8a+4b+2c = 16$

Which has the solution $a = \frac {36} 6 = 6$ , $b = \frac {48 - 12 a}{4} = \frac {48 - 12 \cdot 6}{4} = -6$ , $c = \frac {16 - 8 a - 4 b}{2}= \frac {16 - 8 \cdot 6 - 4 \cdot(-6)}{2} = - 4$

So $f(x) = 6x^3 - 6x^2 - 4x - 16$ and thus $f(5) = 6 \cdot 5^3 - 6 \cdot 5^2 - 4 \cdot 5 - 16 = 564$

RHS is quadratic. Therefore, for LHS to reduce to a quadratic equation, f(x) must be a cubic polynomial. [Note: If you don't understand as to why f(x) would be a cubic polynomial, assume f(x)= a{0} + a{1}x +a{2}x^2 +...+ a{n-1}x^{n-1} +a{n}x^n. Put this in the given equation and you will notice that a{4},a{5},...,a{n} would have to be 0 for the given equation to hold] Therefore, assume a cubic polynomial ax^3 + bx^2 + cx + d =0. (d= -16 because f(0) = -16). Plug in this as f(x) in the given equation, compare the coefficients to get a,b and c. Solving we get a=6, b= -6, c=2. Therefore f(5) = 594

If f(x) is a polynomial of degree n, then g(x)=f(x+2)-f(x) is a polynomial of degree n-1. Since g(x) has degree 2, f(x) must have degree 3. Thus, let f(x) = ax^3+bx^2+cx+d. Then we are given that

g(x) = f(x+2)-f(x) = 6ax^2+(12a+4b)x+(8a+4b+2c) = 36x^2+24x+8

Each coefficient must be equal, so we have three equations in three variables (a,b,c), and solving gives a=6, b=-6, c=-4. We also know that f(0)=-16, so d=-16. Then f(x) = 6x^3-6x^2-4x-16, so f(5) = 564.

Let f(x) be an nth degree polynomial, i.e f(x)=a0 +a1 x +a2 (x^2)+....+an(x^n) (an != 0) .... (1) Given: f(0) = a0 = -16 f(x+2)=a0+a1 (x+2)+a2 {(x+2)^2}+.....+an {(x+2)^n} .... (2) Subtracting (1) from (2) f(x+2)-f(x) = (2 a1+4 a2+8 a3)+{(4 a2+12 a3) x}+(6 a3) (x^2)+.... (3) Comparing (3) with (6x+4)^2 , we get a1= -4; a2= -6; a3= 6 and ak=0 for k>3 Hence, f(x) = 6 (x^3) - 6 (x^2) - 4 x -16 and f(5) = 564

f(x+1)-f(x)=18*x^2+6x-4

Because a finite difference of f has degree 2, f has degree 3. Let f be $ax^3+bx^2+cx+d$.Then $f(x+2)-f(x)=a(x+2)^3+b(x+2)^2+c(x+2)+d-ax^3+bx^2+cx+d=(6x+4)^2$, or $6ax^2+(12a+4b)x+(8a+4b+2c)=36x^2+48x+16$. Solving, we have a = 6, b = -6, c = -4, and because f(0)=-16, d = -16. Plugging 5 in, we have that f(5)=564.

Let me give it a try. This would not be really formal but in a conversational form.

Okay, first of all, the polynomial should be of degree 3. Obviously, it could not be a degree 2, because the different would be a nice linear equations. IT could also not be of degree 4 or above, or else the difference would be at least degree 3.

Now, we know that the polynomial is degree 3. Let's do lagrange!!!

since f(0)=-16, f(2) is -16+(4)^2=0. We also know that f(4) is 0+(12+4)^2=256, and f(6) is 256+(24+4)^2=1040. So we could do lagrange on all of that then we get the polynomial 6x^3-6x^2-4x-16. Plugging in 5, we get 750-150-20-16 which is 564. :-)))

Solution 1: Let $f(x) = a_nx^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots + a_1 x^{1} + a_0$ . Consider the expansion $\begin{array}{c}\ f(x+2) - f(x) & = \left(a_n(x+2)^n + a_{n-1}(x+2)^{n-1} + \ldots \right) - \left(a_nx^n + a_{n-1}x^{n-1} + \ldots \right) & \\ & = a_nx^n +2na_nx^{n-1} + a_{n-1}x^{n-1} - a_nx^n - a_{n-1}x^{n-1} + \ldots & \\ & = 2na_nx^{n-1} + \ldots & \\ \end{array}$

Since this is equal to $36x^2 + 48x + 16$ , we must have $n-1 = 2 \Rightarrow n = 3$ and $2na_n = 36 \Rightarrow a_n = 6$ . Therefore the degree of $f(x)$ must be $n = 3$ with leading coefficient $a_n = 6$ . The constant coefficient is $-16$ since $f(0)=-16$ . Let $f(x) = 6x^3 + ax^2 + bx -16$ . Then $\begin{aligned} f(x+2)-f(x) &= 6(x+2)^3 + a(x+2)^2 + b(x+2) - 16 -(6x^3 + ax^2 + bx - 16) & \\ & = 6 (6x^2+12x+8) + a (4x+4) + b (2) & \\ & = 36x^2 + (72+4a)x + 48 + 4a + 2b & \\ \end{aligned}$

Since this is equal to $36x^2 + 48x + 16$ , we have $a = \frac{48-72}{4} =-6$ and $b = \frac{16-48-4(-6)}{2}=-4$ . Thus $f(x) = 6x^3 -6x^2 -4x-16$ . Hence $f(5) = 564$ .

Solution 2: Similar to the above, we know that $f(x)$ is a cubic. We have $f(-2) = f(0)-8^2 = -16 - 64 = -80$ , $f(0) = -16$ , $f(2) = f(0) + 4^2 = -16 + 16 = 0$ and $f(4) = f(2) + 16^2 = 0 + 256 = 256$ . Hence, by lagrange Interpolation formula,

$f(x) = (-80)\frac {(x-0)(x-2)(x-4)}{(-2-0)(-2-2)(-2-4)} + (-16) \frac {(x+2)(x-2)(x-4)}{(0+2)(0-2)(0-4)} + 0 \frac {(x+2)(x-0)(x-4)}{(2+2)(2-0)(2-4)} + 256 \frac {(x+2)(x-0)(x-2)}{(4+2)(4-0)(4-2)}$

This allows us to evaluate $f(5) = (-80) \frac { (5)(3)(1)}{(-2)(-4)(-6)} + (-16) \frac {(7)(3)(1)}{(2)(-2)(-4)} + 0 + 256 \frac {(7)(5)(3)}{(6)(4)(2)}=564$ .