Question

factor 3a² + 4a + 1; options: (3a - 1)(a - 1), not factorable, (3a + 1)(a + 1), (7a - 10)(a - 8)

Explanation:

Step1: Test the quadratic discriminant

For quadratic $ax^2+bx+c$, discriminant $D=b^2-4ac$.
Here, $3a^2+4a+1$, so $D=4^2-4(3)(1)=16-12=4$

Step2: Check integer factor pairs

Find two numbers that multiply to $3\times1=3$ and add to $4$: $3$ and $1$.

Step3: Factor by grouping

$$\begin{align*} 3a^2+4a+1&=3a^2+3a+a+1\\ &=3a(a+1)+1(a+1)\\ &=(3a+1)(a+1) \end{align*}$$

Answer:

$(3a+1)(a+1)$