I'm having trouble with factorization?

Question

Can you help me with these 2 problems and show me how it is done so I can do the rest myself and leave you all alone :)
Factor completely:
4x(x+6y)+6(x+6y)

y^4+3y^3-5y^2-15y

Answer 1

These are exercises in factoring by grouping. For the first expression, note that both 4x and 6 are both mutiplied by a common binomial factor, (x + 6y). "Both 4x and 6" is written in math as (4x + 6). So we can rewrite this expression as
(4x + 6)(x + 6y). Then you are done. It's factored.

The second expression has 4 terms and is not as far along as the first when it comes to factoring by grouping. The first step is to find the greatest common factor of the first two terms, y^3, and factor this out of these two terms...y^3 (y + 3)

...and then find the greatest common factor of the second two terms, -5y, and factor this from the second two...
-5y(y + 3) and make sure that you ended up with the same binomial in both parentheses (which we did), (y + 3).

Your expression is now y^3 (y +3) -5y (y + 3) and you can now factor this in the same way as the first problem. Both y^3 and -5y are both multiplied by (y+3) so we can rewrite this as (y^3-5y)(y + 3) or y(y^2 -5)(y+3). Done.

Answer 2

First one:

Since you have 4x (x + 6y) + 6(x + 6y), x + 6y is the same in both parts of the equation, so x + 6y is one factor, and the other factor is 4x + 6, so factored completely, it is
(4x + 6) ( x + 6y)

Second one:

y^4 + 3y^3 - 5y^2 - 15y
split this into two parts:
y^4 + 3y^3 - (5y^2 + 15y)
(noticed I factored out a negative sign from the last two terms so you end up with both factors positive for now)

Factor out the GCF from both pairs of terms:

y^3 ( y + 3) - 5y ( y + 3)

When you factor both pairs of terms, you want to end up with the same thing in parentheses, like I did.
The two sets of factors are:
(y^3 - 5y) ( y+3)
(The second terms in parentheses were the terms already outside of parentheses in the last step)

Hope this helps!

Answer 3

This first one prepares us for the second one.

4x(x + 6y) + 6(x + 6y) has two terms and both terms have an element of (x + 6y) so we can factor this out:

(x + 6y)(4x + 6), and we have factored it completely

When you have several elements in a polynomial of degree greater than 3, one approach is to think of it as a couple segments that we try to factor in such a way that they have a common factor:

y^4+3y^3-5y^2-15y = y^2(y^2 + 3y) - 5(y^2 + 3y)
= (y^2 - 5)(y^2 + 3y)
= (y^2 - 5)*y*(y + 3)
= y(y + 3)(y^2 - 5)

--charlie

Answer 4

4x(x+6y) +6(x+6y) Both terms contain (x+6y), so:
(4x+6) (x+6y) The first term can be factored further
2 (2x+3) (x+6y)

y^4+3y^3-5y^2-15y By splitting into 2 parts, it is easier to
(y^4+3y^3) (-5y^2-15y) see how to factor this one
y^3 out of the first; -5y out of the second
y^3(y+3) -5y(y+3) Both terms contain (y+3), so:
(y^3-5y) (y+3) The first term can be factored further
y (y^2-5) (y+3)

Answer 5

4x(x+6y)+6(x+6y)
(4x+6)+(x+6y)


y^4+3y^3-5y^2-15y Common monomial --y
y(y^3+3y^2-5y-15) Factor by grouping
y[y^2(y+3)-5(y+3)] pull out the common binomial
y[(y+3)(y^2-5)] and you get the answer.

Answer 6

4x(x+6y)+6(x+6y)
at first glance , you see that (x+6y) is in the two terms
so (x+6y) (4x+6) and you note that 4 and 6 are even so
2*(x+6y) (2x+3)

y^4+3y^3+5y^2-15y you can put y in factor so

y(y^3+3y^2+5y -15)

Answer 7

the first one :
do begin by multiplying out the 4x and the 6.

4x^2+ 24xy + 6x +36y

then thats it.


idk number 2 sorrrrryyy!