How do you factor 3n^2 - 8n + 4?

2 Answers
May 19, 2015

3n^2-8n+4 = (3n-2)(n-2)

How did I find this?

We are looking for factors in the form (an+b)(cn+d)

(an+b)(cn+d) = acn^2 + (bc+ad)n+bd

Comparing this with our original quadratic we find:
ac = 3
bc+ad = -8
bd = 4

If a, b, c and d are integers then we are looking for integers that satisfy these equations.

ac=3 has possible integer solutions:

a=3, c=1
a=-3, c=-1
a=1, c=3
a=-1, c=-3

We might as well pick the first of these, they will ultimately give very similar factorisations.

Let a=3, c=1

bd > 0. So b and d are either both positive or both negative. If they were both positive, then bc+ad = b+3d would be positive too, which it isn't. So both b and d are negative.

Notice that b+3d = -8 is even. So either b and d are both odd or both even.

So we are looking for two negative whole numbers b and d such that bd = 4 and b and d are both even.

That only leaves the possibility b=d=-2

May 19, 2015

3n^2 - 8n +4

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like an^2 + bn + c, we need to think of 2 numbers such that:

N_1*N_2 = a*c = 3 xx 4 = 12

and

N_1 +N_2 = b = -8
After trying out a few numbers we get N_1 = -2 and N_2 =-6
- 2 xx -6 = 12 and (-2) +(-6)= -8

3n^2 - 8n +4 = 3n^2 - 6n - 2n +4

= 3n(n -2) - 2(n-2)

=color(green)( (3n -2)(n -2))