Math. Factoring out this common factor?

Problem: 3v(v+1) + 1(v+1)

Look at 3v(v + 1) and 1(v+ 1) separately....they both have a (v + 1)

SO, if you take it out...what do you have left? You have (3v + 1)
so your answer is:

Answer: (v+1)(3v+1)



Problem:5y(3y+2) + 2(3y+2)

Look at 5y((3y + 3) and 2(3y + 2) separately....they both
have (3y +2)

SO, if you take out the (3y + 2) from both terms, what do you have left? You have (5y + 2), so your answer is:

Answer:(3y+2)(5y+2)



Problem:y(8y-3) + 2(8y-3)

Same thing here...both terms have a (8y - 3), so if you take it out, you have (y + 2) left...so your answer is:

Answer:(8y-3)(y+2)

hope this helps you
:o)
dr

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e.g.
6 + 14
= 3(2) + 7(2)
= 2(3 + 7) (← factorization)

1)
3v(v + 1) + 1(v + 1)
= (v + 1)(3v + 1)

2)
5y(3y + 2) + 2(3y + 2)
= (3y + 2)(5y + 2)

3)
y(8y - 3) + 2(8y - 3)
= (8y - 3)(y + 2)

Oh my!!! every problem is missing a sign between 2 terms.
Unless that is settled, nothing can be solved.
If you know 3x + 5x = (3+5)x, then 3(y+1) + 5(y+1) = (3+5)(y+1) should not be difficult.
And then all these three are simple after you take care of the middle sign.