[mickno]

There is way too many rules it's all driving me nuts


How would you factorise this



4n2+8n+3

[James A]

(Original post by mickno)
There is way too many rules it's all driving me nuts


How would you factorise this



4n^2+8n+3

I'm presuming the 4n2 means 4n^2

you will notice that the common factor for 'n' appears in the first two terms.

4n2 + 8n + 3

The parts highlighted in bold contain 'n', hence we can only factor the n out there.

Also notice that the highest number that divides both 4 and 8 is 4

so you would be left with

4n(n + 2) + 3

you leave '3' on its own because you can't divide 3 by 4.

EDIT : at the negs? come on people, i left it with the '+3' because i wanted to leave it as integers

[FranticMind]

Its a quadratic equation.

Spoiler:

Show

But the most you could factorise without finding the solutions would be in terms of the n and 4 factor.
n(n+2)+3/4

Using quadratic equation

4n^2+8n+3=0

n^2+2n+3/4

(n+3/2)(n+1/2)=0

[FranticMind]

(Original post by James A)
4n(n + 2) + 3

you leave '3' on its own because you can't divide 3 by 4.

Its nicer as a fraction 3/4. n(n+2)+3/4

[beaver_tron]

calculator does that for me

[electriic_ink]

(Original post by FranticMind)
Its a quadratic equation.

Spoiler:

Show

But the most you could factorise without finding the solutions would be in terms of the n and 4 factor.
n(n+2)+3/4

Using quadratic equation

4n^2+8n+3=0

n^2+2n+3/4

(n+3/2)(n+1/2)=0

OP never said it equalled zero

Spoiler:

Show

So 4n^2+8n+3 = 4(n^2+2n+3/4) = 4(n+3/2)(n+1/2) = (2n+3)(2n+1)

[Education_1]

(Original post by mickno)
There is way too many rules it's all driving me nuts


How would you factorise this



4n2+8n+3

ok so you can do this in 2/3 ways:
1.The ac method
2.The complete the square method if completely desperate (google)
3.USE quadratic formula when nothing else work
EDIT:Always try AC method if this is confusing try complete the square and if both cant seem to work or too complicated to factorise (large numbers); use formula
remember your question is given in for ax2+bx+c (when there is x)
now step one ac observe 4n2+8n+3
1.axc= 4x3=12 (decide 2 numbers that times together to give 12 and add to give 8) so lets start 1x12 gives you 12 but dont add to give 8. lets try another 6x2 is 12 and add to give 8
2. you have your two numbers so write it in this form and factorise it in a way you get two equal set of bracket (you will see what i mean)
4n2+2n+6n+3
factorise

2n(2n+1) 3 (2n+1)
notice how i made it to get 2 common bracket now all you need is
(2n+3) (2n+1)

Hope this help

[FranticMind]

(Original post by Education_1)
ok so you can do this in 2/3 ways:
1.The ac method
2.The complete the square method if completely desperate (google)
3.USE quadratic formula when nothing else work

At the early stage you only use the other methods to teach you basic algebraic manipulation.

The quadratic formula instantly gives you the roots of the equation and is hence a better method. I can use it five or six times in a row just to sort out far more complex equations. Its good practice for when you need it at the higher levels.

[FranticMind]

(Original post by electriic_ink)
OP never said it equalled zero

Spoiler:

Show

So 4n^2+8n+3 = 4(n^2+2n+3/4) = 4(n+3/2)(n+1/2) = (2n+3)(2n+1)

No you set it to zero so you can find the solutions. ie where the curve intercepts the x axis?

[BabyMaths]

(Original post by FranticMind)
No you set it to zero so you can find the solutions. ie where the curve intercepts the x axis?

but as Electriic_ink pointed out the OP's question was about factorising a quadratic expression, not solving an equation.

If you don't see the factorisation easily you could try this:



(Already posted by Education_1, I know, but it's worth repeating)

[zed963]

As the above has stated (2n+1)(2n+3) is the correct answer however there are several methods. Choose the one which you are most comfortable with.

[Alkranite]

whats the ac method?