Factorising

Question

Factorise completely:

a⁴-13a²+3b

Answer 1

I guess 3b was 36.

 

Then we have 

 

a⁴ - 13a²+36=(a²-9)(a²-4)

                   =(a+3)(a-3)(a+2)(a-2)

Answer 2

Substitute y = a² and get a quadratic expression.

 

a⁴ - 13a² + 3b = y² - 13y + 3b = (y + A)(y + B) = (a² + A)(a² + B)

 

where

 

AB = 3b

A + B = -13

 

However, without a value for b, there is not much more we can do.  I am convinced that the 3b is a typo, and that the problem really has 36 instead of 3b, as suggested by Guiyou Q.  Note how that gives a very straightforward factorization.

 

 

 

 

Answer 3

a⁴-13a²+3b

 

 

Look at the first 2 terms:

a⁴ - 13a²

This can be written as the difference of 2 squares:

a⁴ = (a²)²

13a² = [(√13)a]²

 

a⁴ - 13a² + 3b = (a² - (√13)a)(a² +  (√13)a) + 3b

 

I don't know if this is what your teacher is looking for.

The 3b just doesn't fit.