You can factor a polynomial by finding its roots: if [tex]x_0[/tex] is a solution of [tex]p(x)[/tex], i.e. [tex]p(x_0)=0[/tex], then [tex]p(x)[/tex] is divisible by [tex](x-x_0)[/tex].
You keep factoring the polynomial until the remaining factor has no more roots, and is thus irreducible.
In this case, we have
[tex]b^2+3b+2=0 \iff b=\dfrac{-3\pm\sqrt{9-8}}{2} = \dfrac{-3\pm 1}{2} \implies b_1 = -2,\ b_2 = -1[/tex]
So, the polynomial can be written as
[tex]b^2+3b+2=(b+1)(b+2)[/tex]
Answer:
(b + 1)(b + 2)
Step-by-step explanation:
Ask yourself: What are possible factors of the constant, 2? They are {1, 2}.
Do these factors, when added together, result in 3? Yes.
Thus,
b^2+3b+2 = (b + 1)(b + 2) = b^2 + 1b + 2b + 2
