How do you factor #2t^2+7t+3#?

Question

Ethan Adkins · Accepted Answer

Make #2t^(2) + 7t + 3# equal to zero to form a quadratic equation.

Here's the general form of the quadratic equation

#ax^(2) + bx + c =0#

You need to find two numbers that multiply to give #a * c# and that add to give #b#. For this example, the two numbers you need must multply to give #2 * 3 = 6# and add to give #"7"#.

To find them, just list all of the factors of #6# and try to find a pair that satisfies the criteria given

#6 = 1* 2 * 3#

In simple cases such as this one, these two numbers can be easily found to be #"6"# and #"1"#, since

#6 + 1 = 7# and #6 * 1 = 6#

So, your equation then becomes

#2t^(2) + 6t + t +3#

This then becomes

#(2t^(2) + 6t) + (t + 3) => 2t * (t + 3) + (t + 3)#, which can be written as

#(t + 3) * (2t + 1)# #-># this is how the original equation factors.

Eva Abramson · Answer

undefined To factor the quadratic expression 2t^2 + 7t + 3, you need to find two numbers that multiply to give you the product of the coefficient of t^2 (2) times the constant term (3), which is 6, and add up to the coefficient of the linear term (7). These numbers are 6 and 1. Then, rewrite the middle term (7t) using these two numbers. So, the expression can be factored as (2t + 1)(t + 3).