How do you factor \[{{t}^{2}}+8t+12\]?

Hint: If \[x=a\] is a root of a polynomial function, then \[x-a\] is one of its factors. To express a quadratic equation \[a{{x}^{2}}+bx+c\] in its factored form. We have to find its roots, say \[\alpha ,\beta \] are the two real roots of the equation. Then the factored form is \[a\left( x-\alpha \right)\left( x-\beta \right)\]. We can find the roots of the equation using the formula method as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step solution:
We are given the quadratic expression \[{{t}^{2}}+8t+12\]. On comparing the general solution of the quadratic equation \[a{{x}^{2}}+bx+c\], we get \[a=1\].
To express in factored form, we first have to find the roots of the equation \[{{t}^{2}}+8t+12\].
We can find the roots of the equation using the formula method.
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of the coefficients in the above formula, we get
\[\begin{align}
  & \Rightarrow t=\dfrac{-(8)\pm \sqrt{{{\left( 8 \right)}^{2}}-4(1)(12)}}{2(1)} \\
 & \Rightarrow t=\dfrac{-8\pm \sqrt{16}}{2} \\
 & \Rightarrow t=\dfrac{-8\pm 4}{2} \\
\end{align}\]
\[\Rightarrow t=\alpha =\dfrac{-8+4}{2}=-2\] or \[x=\beta =\dfrac{-8-4}{2}=-6\]
Canceling out the common factor from both numerator and denominator for the above fraction, we get
\[\Rightarrow \alpha =-2\] and \[\beta =-6\]
Now that, we have the roots of the given expression, we can express it as its factored form as follows,
For the quadratic expression \[{{t}^{2}}+8t+12\], \[a=1\] and the roots as\[\alpha =-2\And \beta =-6\].
The factored form is,
\[\begin{align}
  & \Rightarrow 1\left( t-\left( -2 \right) \right)\left( t-\left( -6 \right) \right) \\
 & \Rightarrow \left( t+2 \right)\left( t+6 \right) \\
\end{align}\]

Note: As we use the roots of the expression, to make the factored form. We can also use the factored form to find the roots of an expression. It should be noted that an expression can only be expressed as its factored form if it has real roots. For example, the quadratic expression \[{{x}^{2}}+1\] has no real roots. Hence, it can not be expressed as its factored form.