Express $6x^3+35x^2+34x-40$ as a product of linear factors.

 

Try factors of $40$ increasing order, usually a low value of $x$ will work.

Factors of $40$ are $1,2,4,5,,8,10,20,40$ and their negatives.

$P(-1)=-6+35-34-40\ne 0$

$P(1)=6+35+34-40\ne 0$

$P(-2)=-48+140-68-40\ne 0$

$P(-4)=-384+560-136-40=0\ $  !!  yay

Therefore by the factor theorem, $(x-(-4))=(x+4)$ is a factor of $P(x)$.

$\therefore P(x)=(x+4)(6x^2+bx-10)$.

I guessed the quadratics first and last terms as its obvious just looking at it!!

The $x$-term gives us $b$:  $4bx-10x=34x\implies b=11$

$\therefore P(x)=(x+4)(6x^2+11x-10)$.

Factorising the quadratic gives us

$\therefore P(x)=(x+4)(3x-2)(2x-5)$.