[zed963]

I am having trouble factorising this equation. I don't know how to solve this equation. It would be much appreciated if you could explain to me step by step on how to solve this question:

(X+5)+3(x+5)^2

Thanks.

[Pheylan]

If you had instead, would you be able to see what to do?

[elldeegee]

(Original post by zed963)
I am having trouble factorising this equation. I don't know how to solve this equation. It would be much appreciated if you could explain to me step by step on how to solve this question:

(X+5)+3(x+5)^2

Thanks.

either:
expand brackets, simplify, and then factorise again.

or

take out a factor of (x+5)

the latter is what i think the previous poster was getting at,


becomes
and is therefore factorised.

[zed963]

(Original post by elldeegee)
either:
expand brackets, simplify, and then factorise again.

or

take out a factor of (x+5)

the latter is what i think the previous poster was getting at,


becomes
and is therefore factorised.

So we know that a common factor is (x+5) so now I'm left with 3x+15.

[elldeegee]

(Original post by zed963)
So we know that a common factor is (x+5) so now I'm left with 3x+15.

(X+5)+3(x+5)^2
taking out a common factor of (x+5) leaves:

(x+5) (1 + 3(x+5))

See where your mistake would have been when saying you're left with 3x+15?
it's an easy mistake to make

[zed963]

(Original post by elldeegee)
(X+5)+3(x+5)^2
taking out a common factor of (x+5) leaves:

(x+5) (1 + 3(x+5))

See where your mistake would have been when saying you're left with 3x+15?
it's an easy mistake to make

How does the 1 come into play?

[Math12345]

(x+5)+3(x+5)^2

The first thing you should do is make (x+5)^2 into (x+5)(x+5) to make it easier.

(x+5)+3(x+5)(x+5)

The common factor is (x+5) as you can see.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

This is what you should get:

(x+5)(1+3(x+5))
(x+5)(1+3x+15)

[zed963]

If I was to expand this whole number fully would it become x+5 +3x+15 ?

[elldeegee]

(Original post by zed963)
How does the 1 come into play?

say we had the one from before y+3y^2
to factorise this, take out a factor of y:

y(1+ 3y)
because from original and

so in this case,

you've take out a factor of

and

thus,
becomes

simplifying this last bit gives

[TenOfThem]

(Original post by zed963)
If I was to expand this whole number fully would it become x+5 +3x+15 ?

no

[zed963]

(Original post by Math12345)
(x+5)+3(x+5)^2

The first thing you should do is make (x+5)^2 into (x+5)(x+5) to make it easier.

(x+5)+3(x+5)(x+5)

The common factor is (x+5) as you can see.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

This is what you should get:

(x+5)(1+3(x+5))
(x+5)(1+3x+15)

This is the bit that I am not understanding completelym.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

[Math12345]

(Original post by zed963)
This is the bit that I am not understanding completelym.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

(x+5)*y=(x+5)
(x+5)*z=3(x+5)(x+5)

y=1
z=3(x+5)

[TenOfThem]

(Original post by zed963)
This is the bit that I am not understanding completelym.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

Start with some easy examples

Can you factorise

[elldeegee]

(Original post by zed963)
This is the bit that I am not understanding completelym.

What times (x+5) equals (x+5)?
And what times (x+5) equals 3(x+5)(x+5)?

say if we had something like

2x+6

how would you factorise this? you'd say to yourself, what is a factor of both "2x" and "6", or "what goes into "2x" and "6""

The answer would be 2, because 2 is a factor of 2x and 2 is a factor of 6.

so 2x+6 factorised becomes 2(x+3)
and then multiplying this out would give
and

and putting them together gives 2x+6, which is the original before factorisation.


P.s i have a feeling this may be confusing, so much easier to explain this stuff faceto face

[zed963]

(Original post by TenOfThem)
Start with some easy examples

Can you factorise

y(3+5y)

y(3+y)

y(1+5y)

[zed963]

(Original post by elldeegee)
say if we had something like

2x+6

how would you factorise this? you'd say to yourself, what is a factor of both "2x" and "6", or "what goes into "2x" and "6""

The answer would be 2, because 2 is a factor of 2x and 2 is a factor of 6.

so 2x+6 factorised becomes 2(x+3)
and then multiplying this out would give
and

and putting them together gives 2x+6, which is the original before factorisation.


P.s i have a feeling this may be confusing, so much easier to explain this stuff faceto face

I know how to do basic factorising but not when it comes to these complicated questions.

[TenOfThem]

(Original post by zed963)
y(3+5y)

y(3+y)

y(1+5y)

Excellent


So can you do

(anything) + 5(anything)^2

Using that last example of

[zed963]

(Original post by TenOfThem)
Excellent


So can you do

(anything) + 5(anything)^2

Using that last example of

Now thats the bit where I get confused.

[zed963]

I am also having trouble with this thing

(x-y)^2-b(x-y)

[Math12345]

1. Write out the quadratic like this: (x-y)(x-y)-b(x-y)
2. Common factor? (...)
3. Then you collect the other terms in the other bracket (...)