Using the Roots of a Polynomial

Since the only roots of f are 0,2,3, the polynomial $f(x)$ has form: $px^m(x-2)^n(x-3)^l$ , where m,n,l are positive integer.

Therefore, $f'(x)=pmx^{m-1}(x-2)^n(x-3)^l+pnx^m(x-2)^{n-1}(x-3)^l+plx^m(x-2)^n(x-3)^{l-1}$

$=px^{m-1}(x-2)^{n-1}(x-3)^{l-1}[x^2(m+n+l)-x(5m+3n+2l)+6m]$ .

It is obvious that either of 0,2,3 is not root of $8x^2-24x+7$ .

Therefore, in order to make $f'(x)$ be divisible by $8x^2-24x+7$ , $x^2(m+n+l)-x(5m+3n+2l)+6m$ must be divisible by $8x^2-24x+7$

or $x^2(m+n+l)-x(5m+3n+2l)+6m=k(8x^2-24x+7)$ .

Therefore, $m=\frac{7}{6}k$ , $n=\frac{9}{2}k$ , $l=\frac{7}{3}$ .

In order to make m,n,l be positive integer and the sum $m+n+l$ be smallest, we obtain $k=6$ .

The degree of $f(x)$ is: $m+n+l=48$ .

$f(x)= C x^a (x-2)^b (x-3)^c$

$f'(x)=C [a x^{a-1} (x-2)^b (x-3)^c + x^a b(x-2)^{b-1} (x-3)^c + x^a (x-2)^b c(x-3)^{c-1}] =$ $f'(x)=C [a(x-2)(x-3)+bx(x-3)+cx(x-3)] [x^{a-1}(x-2)^{b-1}(x-3)^{c-1}] =$ $f'(x)=C [(a+b+c)x^2+(-5a-3b-2c)x+6a] [x^{a-1}(x-2)^{b-1}(x-3)^{c-1}] =$

$f'(x)$ is divisible by $8x^2-24x+7$ , which means that:

$[(a+b+c)x^2+(-5a-3b-2c)x+6a] = N [8x^2-24x+7]$ or $a+b+c = 8N; -5a-3b-2c=-24N; 6a=7N$

Solution with smallest $a$ is $a =7$ using $N=6$ , resulting in $b=27$ and $c=14$ .

So smallest value for $a+b+c=7+27+14= \fbox {48}$

Since the only roots of $f(x)$ are $0$ , $2$ , $3$ , we can conclude that $f(x)= c x^m (x-2)^n (x-3)^k$ where $c$ is some non-zero constant and $m$ , $n$ , $k$ are non-negative integers. Applying chain-rule of differentiation twice, we can show that: $f'(x)= c x^{m-1} (x-2)^{n-1} (x-3)^{k-1} \{ x(k(x-2)+n(x-3)) + m(x-2)(x-3) \}$

Note that $f'(x)$ will be divisible by $8x^2 - 24x + 7$ iff all roots of $8x^2-24x+7$ are also roots of $f'(x)$ . However, note that the only roots of $c x^{m-1} (x-2)^{n-1} (x-3)^{k-1}$ are $0$ , $2$ , $3$ , none of which are roots of $8x^2-24x+7$ . So the roots of $8x^2-24x+7$ must be the roots of the function in the brackets. We expand the polynomial in the second braces: $x(k(x-2)+n(x-3)) + m(x^2 - 5x + 6)$ $= (m+k+n)x^2 - (2k+3n+5m)x + 6m$ Note that this is also a quadratic polynomial. It is known that if two polynomials of the same degree have all their roots common, then one of them must be a non-zero constant multiplied to the other. Hence, we obtain: $(m+k+n)x^2 - (2k+3n+5m)x + 6m= d(8x^2-24x+7)$ for some non-zero constant $d$ .
Comparing co-efficients, we obtain: $6m= 7d \text{.....(1)}$ $2k+3n+5m= 24d \text{.....(2)}$ $m+k+n= 8d \text{.....(3)}$ We can easily solve these equations for $(m,k,n)$ , and obtain: $m= \frac{7}{6}d$ $n= \frac{9}{2}d$ $k= \frac{7}{3}d$ Note that the degree of $f(x)$ is $m+n+k= 8c$ . Also note that $(m,n,k)$ are positive integers, so $d$ must be a multiple of $lcm(2,3,6)= 6$ . Since we desire the smallest possible value of $8c$ , we set $c=6$ , which gives $(m,n,k)= (7,27,14)$ and the degree of the polynomial= $8 \times 6= \boxed{48}$ .

Let the polynomial be: $f(x)= x^a(x-2)^b(x-3)^c$

Differentiating & taking common we get:

$f'(x)=x^{a-1}(x-2)^{b-1}(x-3)^{c-1}\left(a(x-2)(x-3)+bx(x-3) + cx(x-2)\right)$

$= g(x)*\left(a(x-2)(x-3)+bx(x-3) + cx(x-2)\right)$

Now $(8x^2-24x+7)$ (with say roots $\alpha_1$ & $\alpha_2$ ) doesn't divide $x^{a-1}(x-2)^{b-1}(x-3)^{c-1}$ [this is obvious by factor theorem since $g(\alpha_1),g(\alpha_2) \neq 0$ ]

So we conclude $(8x^2-24x+7)$ is the same quadratic as $(a(x-2)(x-3)+bx(x-3)+cx(x-2)$ or equivalently,

$x^2-3x + \frac{7}{8} = x^2 - \frac{5a+3b+2c}{a+b+c} x + \frac{6a}{a+b+c}$ .

Equating coefficients we arrive at $c=2a$ & $27a=7b \Rightarrow 7|a$ & $27|b$ . To minimize the degree of $f(x)$ ,we must minimize $(a+b+c)$ or equivalently $a,b$ & $c$ . Thus $a=7;b=27;c=14 \Rightarrow (a+b+c)=48$ .

The function must be of the form

$f(x)=a x^b (x-2)^c (x-3)^d$ ,

where $a, b, c$ and $d$ are integers.

By differentiating and collecting terms we get

$f'(x)=ax^{b-1} (x-2)^{c-1} (x-3)^{d-1} \cdot \bigg ( b(x-2)+cx(x-3)+dx(x-2) \bigg )$

$= ax^{b-1} (x-2)^{c-1} (x-3)^{d-1} \cdot \bigg ( (b+c+d)x^2 - (5b+3c+2d)x+6b \bigg )$ .

Since we know that $f'(x)$ is divisible by $8x^2-24x+7$ , this can only be obtained if there exist some numbers so that

$b+c+d=8$

$5b+3c+2d=24$

$6b=7$ .

The solution to this system is $b= \frac{7}{6}$ , $c= \frac{9}{2}$ and $d= \frac{7}{3}$ .

However, our numbers must be integers. We therefore multiply them by $6$ and get

$f(x)=a x^7 (x-2)^{27} (x-3)^{14}$ .

The degree of this polynomial is $7 + 27 +14 = \boxed {48}$ .

Since f(x) has only three roots, f(x) is written in form of Ax^{k}(x-2)^{m}(x-3)^{n} The derivative of f(x) is Ax^{k-1}(x-2)^{m-1}(x-3)^{n-1}*( (k+m+n)x^2 -(5k + 3m+2n)x +6k) The first part x^{k-1}(x-2)^{m-1}(x-3)^{n-1} is indivisible by ( 8x^2 - 24x +7) because x^{k-1}(x-2)^{m-1}(x-3)^{n-1} has only three roots of {2,3,0} and none of them are roots of ( 8x^2 - 24x +7) Therefore the second part ((k+m+n)x^2 -(5k + 3m+2n)x +6k) must be divisible by ( 8x^2 - 24x +7) Or (k+m+n)/8 = (5k+3m+2n)/24 = 6k/7 (k,m,n must be positive integers) For easy calculation I first calculated the system equation: k+m+n= 8; 5k+3m+2n=24; 6k=7 ( this can be done quickly by calculator) Then I got k=\frac{7}{6}; m=\frac{9}{2}; n=\frac{7}{3} To obtain integer, I multiply k,m,n with 6 to obtain k=7;m=24; n=14 Hence, the degree of polynomial f(x) is 7+24+14 = 48

From the given it follows that $f(x)=x^a(x-2)^b(x-3)^c$ for some positive integers $a, b$ and $c$ . We can also easily check that $0, 2$ and $3$ aren't roots of $8x^2-24x+7$ .

Hence $f'(x)=ax^{a-1}(x-2)^b(x-3)^c + x^ab(x-2)^{b-1}(x-3)^c + x^a(x-2)^bc(x-3)^{c-1}$

Rewriting gives;

$f'(x)=x^{a-1}(x-2)^{b-1}(x-3)^{c-1}(a(x-2)(x-3)+bx(x-3)+cx(x-2))$

$=x^{a-1}(x-2)^{b-1}(x-3)^{c-1}((a+b+c)x^2-(5a+3b+2c)x+6a)$

From this we conclude that we must have $8x=a+b+c, 24x=5a+3b+2c$ and $7x=6a$ for some positive integer $x$ .

Solving these for $x$ leads to $a=\frac{7}{6}x, b=\frac{9}{2}, c=\frac{7}{3}x$ . I leave this work to the reader.

Now to solve the problem we need to find the smallest integer $x$ so that $a, b$ and $c$ are positive integers.

Clearly we have $x=6$ , resulting in $a=7, b=27$ and $c=14$ .

The degree of $f(x)$ is therefore $7+27+14=48$ .

Since only 0, 2, and 3 may be the roots, the function f(x) is of the form $m(2-x)^a*(3-x)^b*x^c$ , where m is a constant. The m can be dropped for convenience, because it does not affect the degree of the polynomial. The derivative of this can be found using the chain and product rules. However, as (2-x),(3-x), and x are not factors of $8x^2-24x+7$ all such terms that are factors of the derivative can be dropped. In other words, factorable elements of the derivative can be ignored.

This leaves one with only $x*a*(x-3)+x*b*(x-2)+c*(x-2)(x-3))$ . This can be expanded to $ax^2+bx^2+cx^2-3ax-2bx-5cx+6c$ . For this to have a constant that is a multiple of 7, c, as in the only constant term, 6c, must be at least 7. For the sake of minimization, that value will be used. Since the 7 is multiplied by 6, for the expression to be divisible by $8x^2-24x+7$ , this entire expression must be divisible by 6. The sum of the final coefficients of $x^2$ in the expression in the derivative should be 8 (From the coefficient of $x^2$ in $8x^2-24x+7)$ , since they should match for divisibility) multiplied by 6, due to the necessity for 6 to be factorable from this entire portion, or 48. Since the coefficients of $x^2$ are a, b, and c, and it has been shown that a+b+c=48, and the sum of the powers of the linear elements is the total degree, 48 is the smallest degree of a polynomial that meets the criteria presented.

Since the only roots of $f(x)4$ are $0$ , $2$ , and $3$ , $f(x)$ must be of the form $f(x)=ax^{k}(x-2)^{l}(x-3)^{m}$

By the product rule, $f'(x)=a(kx^{k-1}(x-2)^{l}(x-3)^{m}+lx^{k}(x-2)^{l-1}(x-3)^{m}+mx^{k}(x-2)^{l}(x-3)^{m-1})$ $=ax^{k-1}(x-2)^{l-1}(x-3)^{m-1}(k(x-2)(x-3)+lx(x-3)+mx(x-2))$ $0$ , $2$ , and $3$ are not roots of $8x^{2}-24x+7$ , and the constant $a$ won't help because the coefficients $8$ , $-24$ , and $7$ are relatively prime, so that must divide $k(x-2)(x-3)+lx(x-3)+mx(x-2)$ . This expression is equal to $(k+l+m)x^{2}-(3l+2m+5k)x+6k$ The least value of $k$ for which this is possible is $7$ , so we set $k=7$ and see if $l$ and $m$ work out to be integers. Since $k=7$ , we have $(k+l+m)x^{2}-(3l+2m+5k)x+6k=56x^{2}-144x+42$ So $l+m=41$ and $3l++2m=109$ . This system has solution $l=27, m=14$ , so $k=7$ works. The degree of $f(x)$ is $k+l+m=48$ .

Since all roots of f(x) are in the set {0,2,3}, f(x) must be of the form: $$f(x)=nx^k(x-2)^l(x-3)^m \qquad (1)$$ Its derivative is: $$f'(x)=n\Big(kx^{k-1}(x-2)^l(x-3)^m + x^k l (x-2)^{l-1}(x-3)^m + x^k(x-2)^l m(x-3)^{m-1}\Big)$$ $$f'(x)=n\Big(k(x-2)(x-3) + l x(x-3) + m x(x-2)\Big)x^{k-1}(x-2)^{l-1}(x-3)^{m-1}$$ $$f'(x)=n\Big((k+l+m)x^2 - (5k+3l+2m)x+6k\Big)x^{k-1}(x-2)^{l-1}(x-3)^{m-1}$$

Since none of the roots of $8x^2-24x+7$ is either 0, 2, or 3, it follows that $$n\Big((k+l+m)x^2 - (5k+3l+2m)x+6k\Big)$$ must be divisible by $8x^2-24x+7$ . Since the degrees are the same, we might as well set them equal to each other. Solving the system, where we pick n such that the solution has the smallest integers, yields $k=7, l=27, m=14$ with $n=1/6$ .

From (1) we can see that the smallest degree of f(x) is $k+l+m=7+27+14=48$ . $\qquad \blacksquare$