$ a= \frac{2}{5}; f(x) = 10x^3 +6x^2 + x -2 $
Have difficulties with advanced polynomials of this type. I can not factor.To find the right root. Here is my solution:
$f(x) = 10x^3 +6x^2 + x -2 = 10x^3 +6x^2 +1x^2 -1x^2 +x -2 = (5x^2+x)(2x+1) -x^2 -2 $
I do not understand what I have to do.
On
If you want to show that $a$ is a root of $f$, just plug in $a=2/5$ for $x$ in the definition of $f$. That is to say, find $f(2/5)$.
If $f(2/5) = 0$, then $2/5$ is a root of $f$. Similarly, if it's not zero, then it's not a root.
If you want know how you can conclude that $2/5$ might be a root, the rational root theorem might be worth looking at. You still have to test the potential roots the theorem gives you, but it's better than trying to mess with that cubic right off the bat.
$f(\dfrac25)=\dfrac {80}{125}+\dfrac {24}{25}+\dfrac25-2=\dfrac{80}{125}+\dfrac{120}{125}+\dfrac{50}{125}-\dfrac{250}{125}=0$. Thus it is a root.