Factor the expression by grouping. First, the expression needs to be rewritten as $2t^{2}+at+bt+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=2\times 1=2$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=2$$
Rewrite $2t^{2}+3t+1$ as $\left(2t^{2}+t\right)+\left(2t+1\right)$.
$$\left(2t^{2}+t\right)+\left(2t+1\right)$$
Factor out $t$ in $2t^{2}+t$.
$$t\left(2t+1\right)+2t+1$$
Factor out common term $2t+1$ by using distributive property.
$$\left(2t+1\right)\left(t+1\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$2t^{2}+3t+1=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$t=\frac{-3±\sqrt{3^{2}-4\times 2}}{2\times 2}$$
Square $3$.
$$t=\frac{-3±\sqrt{9-4\times 2}}{2\times 2}$$
Multiply $-4$ times $2$.
$$t=\frac{-3±\sqrt{9-8}}{2\times 2}$$
Add $9$ to $-8$.
$$t=\frac{-3±\sqrt{1}}{2\times 2}$$
Take the square root of $1$.
$$t=\frac{-3±1}{2\times 2}$$
Multiply $2$ times $2$.
$$t=\frac{-3±1}{4}$$
Now solve the equation $t=\frac{-3±1}{4}$ when $±$ is plus. Add $-3$ to $1$.
$$t=-\frac{2}{4}$$
Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.
$$t=-\frac{1}{2}$$
Now solve the equation $t=\frac{-3±1}{4}$ when $±$ is minus. Subtract $1$ from $-3$.
$$t=-\frac{4}{4}$$
Divide $-4$ by $4$.
$$t=-1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{1}{2}$ for $x_{1}$ and $-1$ for $x_{2}$.
