Question

Can `y=x^3+8x^2+17x+10` be factored? If so what are the factors ?

Answer 1

Yes, it can.
`y=(x+1)(x+2)(x+5)`

Explanation:

To factorize a n-degree-polynomial function, one must discover its roots that will allow to rearrange the function in this way
`f(x)=(x-x_1)(x-x_2)[...]"(x-x_n)`

For a polynomial of the third degree, when all coefficients are real numbers, there's at least one real root. The other two roots are either real numbers or complex conjugate numbers.

For a general solution of roots of the third degree equation, requiring some ability to work with complex numbers, I recommend this source:
Roots of a cubic function

But we can try to resolve the problem the easy way.

The coefficient d may help us to find at least one real root.
`d=18=2*5`
This is why we should try the possible roots `+-1, +-2, +-10` and also `+-1/2, +-1/5 and +-1/10`#.

Luckily `x_1=-1` is a root of the function:
`f(x=-1)=-1+8-17+10=0`

Then we should divide the polynomial by `(x-x_1)=(x+1)`:
`" "x^3+8x^2+17x+10" "`|`" "x+1`
`-x^3-x^2" "`|____
___#" "x^2+7x+10#
`" "7x^2+17x`
`" "-7x^2 -7x`
`" "` __
`" "10x+10`
`" "-10x-10`
`" "` _____
`" "0`

The second degree factor can still be further factorized:
`x^2+7x+10=0`
`Delta=49-40=9`
`x=(-7+-sqrt(9))/2` => `x_2=-2`; `x_3=-5`

Then the function factorized is
`f(x)=(x-x_1)(x-x_2)(x-x_3)=(x-1)(x-2)(x-5)`