alright,
1.) since there is a 2 in front of the x^2, you have to multiply 2×21 and you would get 42.
2.) Now that you have 42, find two numbers that multiply to 42 but at the same time, they add to 17. The two numbers that do this are 14 and 3 because 14*3=42 and 14+3=17.
3.) Bring back the 2x^2 and replace the original number 17 with +14x+3x and then bring back the 21. the problem should look like 2x^2+14x+3x+21.
4.) find common factors for each side, i.e find factors for 2x^2+14x and 3x+21. For 2x^2+14x, a 2x can be factored from both because 2 goes into itself once and 14,
seven times, and both have an x. This would give you 2x (x+7). For 3x+21, a 3 can be factored out because 3 goes into itself once and 21 seven times. this would give you 3(x+7). your problem should look like 2x(x+7)3(x+7).
5.) as you can see, there are two (x+7)'s. this is half of your answer. the other half would be your factors 2x and 3, which put together would look like (2x+3). Your answer is (x+7)(2x+3)
1.) 2x^2+17+21
2×21=42
2.) 14 & 3
3.) 2x^2+14x+3x+21
4.) 2x(x+7)3(x+7)
5.) =(x+7)(2x+3)
hope I helped! :)
