What is the difficulty in factoring a trinomial with a leading coefficient that is not to 1?

Answer 1

To factorise a trinomial (or Quadratic) expression, a different approach must be taken than the usual "what adds to the x-coefficient and multiplies to the constant".
As an example, let's use 4x2+ 10x + 4.

First, we check to see if the coefficients all have a common factor.

In this case, all of them can be divided by 2, so our expression becomes 2(2x2+ 5x + 2).

Next, we multiply the x2coefficient by the constant to get a "synthetic" constant. The constant is still 2, but we need to use this new, synthetic constant to continue.

2*2=4, so now we have 2(2x2+ 5x + 4).

Now, we find two numbers that add to 5 and multiply to this new synthetic constant 4: they are 4 and 1.

Now, we split the middle 5x term into two terms with coefficients 4 and 1, and we also bring back our old friend 2 as the original constant. This gives us:

2(2x2+ 4x + 1x + 2)

Next, we factorise the first and second terms together, and the third and fourth together. This gives us:

2( 2x(x+2) + 1(x+2) )

Then we tidy that up by putting the '2x' and the '1' into their own bracket, keeping it next to the (x+2) bracket to get:

2((2x+1)(x+2))

Remove the outermost set of brackets, as they serve no more purpose, to leave us with the final, fully factorised result:

4x2+ 10x + 4 = 2(2x+1)(x+2).