factor: (6ab+4a)+(3b+2)?

Question

your answer should be in the form 2a(?????)+(?????)

Answer 1

(6ab+4a)+(3b+2)
= 2a(3b + 2) + 1(3b + 2)
= (2a + 1)(3b + 2)

Answer 2

First: factor the first set > find the greatest common factor divisble by 6ab and 4a which is 2a.

2a(3b + 2) + 1(3b + 2)

(3b + 2)(2a - 1)

Answer 3

=2a(3b+2)+(3b+2). just take 2a common from 6ab+4a and if you feel easy you can also take 1 common from 3b+2 for doing the question easily.

Answer 4

6 is the product of 3 and 2

6*a*b = 3*2*a*b

4 is the product of 2 and 2

4*a = 2*2*a

Thus; 6ab+4a = 3*2*a*b + 2*2*a
Take 2*a common from both the terms

6ab+4a = 2*a(3*b + 2)

Therefore (6ab+4a)+(3b+2) = 2a(3b+2) + (3b+2)

Answer 5

2a(3b+2) + (3b+2)

When you take out (3b+2) from the first term, you are left with 2a. With the second term, you have to put a placeholder of 1 because you are taking out all of (3b+2).

So you get (2a+1) (3b+2)

Answer 6

(6ab+4a)+(3b+2)

Ans = (2a+1)(3b+2)

or 2a(3b+2) + (3b+2)

Answer 7

(6ab+4a)+(3b+2)
(2a)(3b+2)+(3b+2)

Also:
(2a)(3b+2)+(3b+2)
(3b+2)(2a+1)

Answer 8

i think of you're asking to component the polynomial. the coefficient of the a^2 term has factors 2*2 or a million*4 and coeff of b^2 term merely has a million*3 with a (-) on one in each and every of the two. we want a mixture of them that provides the middle coefficient, -4. you in basic terms wager and attempt and locate that 2*(-3) + 2 * a million = -4 so the factored poly is (2a +b ) * (2a - 3b)

Answer 9

2a(3b+2)+(3b+2)

or you can write it as
(2a+1)(3b+2)

Answer 10

(2a+1)(3b+2).