Orestis_Lignos wrote: (a) Find the value of the real number $k$, for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients. $2^3-2k+2=0$ $\implies$ $\boxed{k=5}$ and then, just dividing by $x-2$ : $\boxed{x^3-5x+2=(x-2)(x^2+2x-1)}$

Orestis_Lignos wrote: (b) If the positive real numbers $a,b$ satisfy the equation $$2a+b+\frac{4}{ab}=10,$$find the maximum possible value of $a$. We have $5-a=\frac b2+\frac 2{ab}>0$ and so $a<5$ Squaring, we have $(5-a)^2=\frac{b^2}4+\frac 2a+\frac 4{a^2b^2}$ $=\frac 4a+\left(\frac b2-\frac 2{ab}\right)^2\ge \frac 4a$ And so $a(a-5)^2\ge 4$, which is $(a-4)(a^2-6a+1)\ge 0$, which, associated to $a<5$, implies $\boxed{a\le 4}$, which indeed fits (with $b=1$)

pco wrote: Orestis_Lignos wrote: (b) If the positive real numbers $a,b$ satisfy the equation $$2a+b+\frac{4}{ab}=10,$$find the maximum possible value of $a$. $(a-4)(a^2-6a+1)\ge 0$, which, associated to $a<5$, implies $\boxed{a\le 4}$, which indeed fits (with $b=1$) Why cannot be $3+\sqrt 2 \leqslant a < 5$ ? EDIT: oops.. I made a mistake when I did the expansion of the $a^2 -6a+1$.

For part b Quote: b) If the positive real numbers $a,b$ satisfy the equation$$2a+b+\frac{4}{ab}=10$$ $$2a=10-\left(b+\frac{4}{ab}\right)\leq 10-4a^{-1/2}$$hence for max value $$2a-10+4a^{-1/2}=0$$, which yields the max value as $4$.

Slave wrote: Why cannot be $3+\sqrt 2 \leqslant a < 5$ ? Because for $3+\sqrt 2 \leqslant a < 5$, we have $a-4>0$ and $a^2-6a+1<0$ and so $(a-4)(a^2-6a+1)<0$; not suitable

a) As 2 is root of $P(x)$, $2^3 - 2k + 2 = 0$ $<=>$ $10 - 2k = 0$ $<=>$ $-2k = -10$ $<=>$ $2k = 10$ $<=>$ $k = 5$ Now, our polynomial $P(x)$ states $x^3 - 5x + 2$. Since 2 is a root, we shall divide $P(x)$ with $2$, we will use long division. $(x^3 - 5x + 2) : (x - 2) = x^2 + 2x - 1$. b) As $b + \frac{4}{ab}$ is positive, $10 - 2a$ is also positive. That means $a < 5$ Similar calculations give us $b < 10$ For $b = 1$, the equation becomes $2a + \frac{4}{a} = 9$ and it's solutions is $a = 4$. For $b > 1$, $a$ gets smaller and smaller. For $b < 1$, $a$ either gives us $a > 5$ or $a < 4$. Hence the maxiumum value od $a$ is $4$