Make 2t^(2) + 7t + 32t2+7t+3 equal to zero to form a quadratic equation.
Here's the general form of the quadratic equation
ax^(2) + bx + c =0ax2+bx+c=0
You need to find two numbers that multiply to give a * ca⋅c and that add to give bb. For this example, the two numbers you need must multply to give 2 * 3 = 62⋅3=6 and add to give "7"7.
To find them, just list all of the factors of 66 and try to find a pair that satisfies the criteria given
6 = 1* 2 * 36=1⋅2⋅3
In simple cases such as this one, these two numbers can be easily found to be "6"6 and "1"1, since
6 + 1 = 76+1=7 and 6 * 1 = 66⋅1=6
So, your equation then becomes
2t^(2) + 6t + t +32t2+6t+t+3
This then becomes
(2t^(2) + 6t) + (t + 3) => 2t * (t + 3) + (t + 3)(2t2+6t)+(t+3)⇒2t⋅(t+3)+(t+3), which can be written as
(t + 3) * (2t + 1)(t+3)⋅(2t+1) ->→ this is how the original equation factors.
