In order to factor a polynomial [tex]p(x)[/tex], we have to find all its roots [tex]x_1,\ x_2,\ldots x_n[/tex] so that we can rewrite the polynomial as[tex]p(x)=(x-x_1)(x-x_2)\ldots(x-x_n)[/tex]This is exactly the same idea we apply when we factor numbers: we look for all the primes that divide the number, and then we write[tex]n = p_1^{e_1}\cdot p_2^{e_2}\ldots p_n^{e_n}[/tex]When talking about polynomials, the idea of prime numbers is represented by irreducible polynomials, i.e. polynomials with no roots.So, we have to find a root of our polynomial. Using the rational root theorem, we can check that [tex]x=-1[/tex] is a solution:[tex]p(-1)=2(-1)^3+9(-1)^2+10(-1)+3 = -2+9-10+3=0[/tex]So, our polynomial is divisible by [tex](x+1)[/tex]. The long division yields[tex]\dfrac{2x^3+9x^2+10x+3}{x+1} = 2x^2+7x+3[/tex]Which is the same as[tex]2x^3+9x^2+10x+3=(x+1)(2x^2+7x+3)[/tex]We can complete the factorization by breaking the quadratic equation: using the standard quadratic formula we can find the solutions[tex]2x^2+7x+3=0 \iff x=-\dfrac{1}{3}\text{ or }x=-3[/tex]Which implies[tex]2x^2+7x+3=\left(x+\dfrac{1}{2}\right)(x+3)[/tex]And finally[tex]2x^3+9x^2+10x+3=(x+1)\left(x+\dfrac{1}{2}\right)(x+3)[/tex]