Question

I'm supposed to apply the method of grouping to a quadratic trinomial (second degree trinomial).

This is the trinomial that I chose:

x2 + 10x + 20

This is what I've done so far:

To apply the same grouping method to factor this quadratic trinomial, it needs to be expanded to at least 4 terms, so that there can be at least 2 terms on each side. When it is expanded, it becomes:

x2 + 5x + 5x + 20

Then, just split the four-term polynomial into 2 groups. It will then look like this:

(x2 + 5x) + (5x + 20)

Answer 1

x can be taken out from the first binomial set. x2 divided by x equals
x, and 5x divided by x simply equals 5. So when x is divided from both terms, it will look like this:

x2(x + 5) + (5x + 20)

5 can be taken out from the second binomial set. 5x2 divided by 5 equals x2, and 20 divided by 5 equals 4. It would then look like this

x2(x + 5) + 5(x + 4)

x2(x + 6) + 5(x + 4)

Answer 2

@triciaal @mathmale @mathstudent55 @zepdrix

Answer 3

The problem is, I don't know how to simplify any further, and for some reasons the calculators are giving a completely different answer than what I got.

Answer 4

Oh and don't worry about the exponents, I just copied this off my word document, the exponents are on there.

Answer 5

You cannot come up with a random quadratic trinomial and expect that it can be factored. Most trinomials are not factorable.

Answer 6

why did you chose that? it is not easily factorable
need factors of the product (a*c) that add up to (b) to make it factorable
otherwise need to use the quadratic formula

suggest change to 9x

Answer 7

9x?

Answer 8

Alright

Answer 9

If you want to make sure you have a factorable trinomial, start by multiplying to binomials together. Then you know the result is definitely factorable.

Answer 10

In this case, with a middle term of 10x it is not factorable, but with a middle term of 9x it is factorable. @triciaal is getting you one the right track.