Question

Factorise 2x^3+9x^2+10x+3​

Answer 1

In order to factor a polynomial $p(x)$, we have to find all its roots $x_1,\ x_2,\ldots x_n$ so that we can rewrite the polynomial as

$p(x)=(x-x_1)(x-x_2)\ldots(x-x_n)$

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

$n = p_1^{e_1}\cdot p_2^{e_2}\ldots p_n^{e_n}$

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 $x=-1$ is a solution:

$p(-1)=2(-1)^3+9(-1)^2+10(-1)+3 = -2+9-10+3=0$

So, our polynomial is divisible by $(x+1)$. The long division yields

$\dfrac{2x^3+9x^2+10x+3​}{x+1} = 2x^2+7x+3$

Which is the same as

$2x^3+9x^2+10x+3=(x+1)(2x^2+7x+3)$

We can complete the factorization by breaking the quadratic equation: using the standard quadratic formula we can find the solutions

$2x^2+7x+3=0 \iff x=-\dfrac{1}{3}\text{ or }x=-3$

Which implies

$2x^2+7x+3=\left(x+\dfrac{1}{2}\right)(x+3)$

And finally

$2x^3+9x^2+10x+3=(x+1)\left(x+\dfrac{1}{2}\right)(x+3)$