factor the following prob and explain plz?

(2w+7)(w+3)

To understand how, multiply it out.

(2w+7)(w+3) are the factors of 2w^2+13w+21

Steps:

1.) Think of a factor of 2w^2 and enclose it in parentheses. I came up with 2w and w to have2w^2. (2w )(w ).

2.) Put a plus sign next to the first factors because there isn't any minus sign in the problem. (2w+ )(w+ ).

3.) Think of the factors of the last term which is 21. It's only 7 and three, and try to put in the parentheses. It's trial and error so you must try to solve first before you come up with the final answer which is (2w+7)(w+3).

To check if correct:

Use the Foil Method or Parrot Beak method and multiply.

Foil Method:

First, Outer Inner, Last

F: 2w x w= 2w^2
O: 2w x 3= 6w
I: 7 x w= 7w
L: 7 x 3= 21

And you should combine the like terms and come up with: 2w^2+6w+7w+21 and simplified as 2w^2+13w+21.

Hope It helps.

you must find two factors whose product is the product of the constant and the numerical coefficient of the first term which is 42 and the sum of the factors which will be the coefficient of the middle term 13

the two factors are 6 and 7

then spilt 13 into 6 and 7

2w^2 + 6w + 7w + 21
factoring by grouping
2w(w + 3) + 7(w + 3)
(2w + 7)(w+3)

2W^2+13W+21
= 2W^2 + (6W+7W) + 21
= (2W^2 + 6W)+(7W + 21)
= 2W(W + 3)+7(W + 3)
= (W+3)(2W + 7)

2w^2 + 13w + 21
= (2w + 7)(w + 3)

2w^2 + 13w + 21 = 2w^2 + 6w +7w + 21 = 2w(w+3) +7(w+3)=(w+7)(2w+7). Does this work for you?

(2w+7)(w+3)