gavrilos wrote: Let $P(x)=ax^3+(b-a)x^2-(c+b)x+c$ and $Q(x)=x^4+(b-1)x^3+(a-b)x^2-(c+a)x+c$ be polynomials of $x$ with $a,b,c$ non-zero real numbers and $b>0$.If $P(x)$ has three distinct real roots $x_0,x_1,x_2$ which are also roots of $Q(x)$ then: A)Prove that $abc>28$, Writing $aQ(x)=(x-u)P(x)$ for some $u$, we get : Identifying constant terms in this equation, we get $u=-a$ Identifying $x$ coefficients, we get $b-a=\frac ca$ Identifying $x^2$ coefficients, we get $b-a=\frac{b+c}{2a}$ Identifying $x^3$ coefficients, we get $b-a=\frac ba$ So $a\ne 1$ and $b=c=\frac {a^2}{a-1}$ and the constraint $b>0$ implies $a>1$ Then $abc=\frac{a^5}{(a-1)^2}$ whose minimum when $a>1$ is $\frac{3125}{108}>28$ QED

gavrilos wrote: B)If $a,b,c$ are non-zero integers with $b>0$,find all their possible values. From $b=\frac{a^2}{a-1}=a+1+\frac 1{a-1}$ positive integer, we get $a=2$ and so $\boxed{(a,b,c)=(2,4,4)}$ Then, we get $P(x)=2x^3+2x^2-8x+4=2(x-1)(x^2+2x-2)$ has indeed three distinct real roots which are also roots of $Q(x)=x^4+3x^3-2x^2-6x+4=(x+2)(x-1)(x^2+2x-2)$

Please help me understand how the minimum calculated for a^5/(a-1)^2 as 3125/108. Is it 32, when a= 2; 2^5/(2-1)^2 = 32

madh1 wrote: Please help me understand how the minimum calculated for a^5/(a-1)^2 as 3125/108. Is it 32, when a= 2; 2^5/(2-1)^2 = 32 $f(x)=\frac{x^5}{(x-1)^2}$ $\implies$ $f'(x)=\frac{(3x-5)x^4}{(x-1)^2}$ and so local minimum over $(1,+\infty)$ when $x=\frac 53$ And $f(\frac 53)=\frac{5^5}{3^32^2}=\frac{3125}{108}$

f'(x) denominator shouldn't it be (x-1)^3

madh1 wrote: f'(x) denominator shouldn't it be (x-1)^3 Yep, just typo. It doest not change anything to the result.

pco wrote: gavrilos wrote: Let $P(x)=ax^3+(b-a)x^2-(c+b)x+c$ and $Q(x)=x^4+(b-1)x^3+(a-b)x^2-(c+a)x+c$ be polynomials of $x$ with $a,b,c$ non-zero real numbers and $b>0$.If $P(x)$ has three distinct real roots $x_0,x_1,x_2$ which are also roots of $Q(x)$ then: A)Prove that $abc>28$, Writing $aQ(x)=(x-u)P(x)$ for some $u$, we get : Identifying constant terms in this equation, we get $u=-a$ Identifying $x$ coefficients, we get $b-a=\frac ca$ Identifying $x^2$ coefficients, we get $b-a=\frac{b+c}{2a}$ Identifying $x^3$ coefficients, we get $b-a=\frac ba$ So $a\ne 1$ and $b=c=\frac {a^2}{a-1}$ and the constraint $b>0$ implies $a>1$ Then $abc=\frac{a^5}{(a-1)^2}$ whose minimum when $a>1$ is $\frac{3125}{108}>28$ QED why is aQ(x) why not just Q(x)