Question

~~Fan + Medal~~
Can someone please explain how a four-term polynomial is factored by grouping and when a quadratic trinomial can be factored using this method. Include examples in your explanation.

Answer 1

For a four-term polynomial factorization:
\[x^3+2x^2-x-2=0\]
\[(x^3-x)+(2x^2-2)=0\]this is the step of grouping, then:

take common factors:
\[x(x^2-1)+2(x^2-1)=0\]

take \[(x^2-1)\] as a common;
\[(x^2-1)(x+2)=0\]

then expand \[(x^2-1)\]:
\[(x-1)(x+1)(x+2)=0\]

Answer 2

Thank you!

Answer 3

What if (x^2-1) ain't a common factor

Answer 4

Excuse me, what is "quadratic trinomial can be factored using this method"?

Answer 5

Um I'm not sure

Answer 6

"What if (x^2-1) ain't a common factor"
you search for the best composition in order not to be in that situation.
If you find that there is no common, then you can complete a quadratic ,,,,, another way

Answer 7

"Um I'm not sure"

how can I solve such thing you don't sure about ;;)

Answer 8

It just what the question asked, I'm searching in the lesson for the answer

Answer 9

I figured it out a quadratic trinomial is just x^2 + bx + c and they want it to be solved by factoring by grouping.

Answer 10

HOW "solved by factoring by grouping."?

Answer 11

1. Check for GCF
2. Split the middle term
3. Factor by grouping
4. Check by distributing