+0

# Cubic Polynomial. Help thanks.

0
49
2
+18

All of the roots of the cubic polynomial $$f(x) = 3x^3+bx^2+cx+d$$ are negative real numbers. If f(0) =81, find the smallest possible value of f(1). Prove that your answer is the minimum.

Jun 4, 2021

#1
+131
0

Here is my take on this question, I could be wrong (inequalities are tricky for me).

From the statement f(0)=81, we have d=81

This, f(x) becomes 3x^3+bx^2+cx+81.

Let the roots be r, s, and t.  Thus, the product rst is equal to -27 by Vieta's formulas.

b=-(r+s+t), which is a positive number, given that r, s, and t are all negative.  Hence, to minimize b, we want to maximize r+s+t, knowing that they are negative.  We seek to relate rst to r+s+t through inequality, and the AM-GM works for this.  The problem arises though, that the AM-GM only works for positive numbers, and r, s, and t are each negative.  Luckily, by setting -r, -s, and -t into the equation, we have all positive inputs, hence allowing for the AM-GM to come into use.

The AM-GM for 3 variables gives $$\frac{-r-s-t}{3}≥ \sqrt[3]{-rst}$$, simplifying to $$-(r+s+t)≥ 3\sqrt[3]{27}$$ or $$r+s+t≤-9$$ from this, the maximum value of r+s+t is -9, so the minimum value for b is 9.  This can only be achieved when r=s=t however, so r=s=t=-3

Now to find the minimum value for c

With r, s, and t each being negative rs, st, and rt must all be positive.  From this, it is easy to apply AM-GM for 3 variables

$$\frac{rs+st+rt}{3}≥\sqrt[3]{rst^2}$$, simplifying to $$rs+st+rt≥27$$.  Equality only occurs when the equality condition of AM=GM is met, however, or when rs=st=rt, r=s=t.  From this, the minimum value of c is 27.

Combining the minimum value of b with the minimum value of c, we find that both can be achieved only when r=s=t=-3.  Through this, the roots must be -3, -3, and -3, with b=9, c=27.  F(1) is equal to 3+b+c+81, or 3+9+27+81, or 120.

Hopefully, this is a valid solution, have a great day!

Note:  After expanding (x+3)^3, and multiplying by 3, I found that this solution is incorrect.  Hopefully, you can find inspirational value from this then.  Where I went wrong was when I tried to use AM-GM to find b, as the application was completely out of the restriction of the AM-GM.

Someone previously asked this question on this website though, and the answer they got was 204.

Jun 4, 2021
edited by EnchantedLava68  Jun 4, 2021
#2
+131
0

After reviewing my work, the next day after this question I found my mistake was not multiplying b and c by 3.  This was because f(x) had roots of -3, -3, -3, but it was multiplied by 3, hence giving -b/3=-9, and c/=27

From this, b=27 and c=81, so f(1) = 3+27+81+81 = 162+30=192

EnchantedLava68  Jun 5, 2021