How do you factor 3n2−8n+4?

2 Answers
George C.
May 19, 2015

3n2−8n+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)=acn2+(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

Don't Memorise
May 19, 2015

3n2−8n+4

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like an2+bn+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=3×4=12

and

N1+N2=b=−8
After trying out a few numbers we get N1=−2 and N2=−6
−2×−6=12 and (−2)+(−6)=−8

3n2−8n+4=3n2−6n−2n+4

=3n(n−2)−2(n−2)

=(3n−2)(n−2)

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