Why does factoring polynomials by grouping work?

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Why does factoring polynomials by grouping work?

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Answer 1 · 2015-03-12T00:42:07

It works for some polynomials but not for others. Mostly , it works for this polynomial because the teacher, or author, or test-maker, chose a polynomial that could be factored this way.

Example 1

Factor: $3 x^{3} + 6 x^{3} - 5 x - 10$ $3x^3+6x^3-5x-10$

I group the first two terms and take out any common factor of those two:
$(3 x^{3} + 6 x^{2}) - 5 x - 10 = 3 x^{2} (x + 2) - 5 x - 10$ $(3x^3+6x^2)-5x-10=3x^2(x+2) -5x-10$

Now I'll take out any common factors in the other two terms. If i get a monomial times $(x + 2)$ $(x+2)$ then factoring by grouping will work. If I get something else, it won't work.

Ther common factor of $(- 5 x - 10)$ $(-5x-10)$ is $- 5$ $-5$ . Taking that factor out leaves $- 5 (x + 2)$ $-5(x+2)$ so we know factoring by grouping will work.

$3 x^{3} + 6 x^{2} - 5 x - 10 = (3 x^{3} + 6 x^{2}) + (- 5 x - 10)$ $3x^3+6x^2-5x-10 = (3x^3+6x^2)+(-5x-10)$

$= 3 x^{2} (x + 2) - 5 (x + 2)$ $=3x^2(x+2)-5(x+2)$ .

Now we have two terms with a common factor $C$ $C$ where $C = (x - 2)$ $C=(x-2)$ . So we have $3 x^{2} C - 5 C = (3 x - 5) C$ $3x^2C-5C=(3x-5)C$

That is: we have $(3 x^{2} - 5) (x + 2)$ $(3x^2-5)(x+2)$

We'll stop there if we're only willing to use integer (or rational) coefficients.

Example 2

Factor: $4 x^{3} - 10 x^{2} + 3 x + 15$ $4x^3-10x^2+3x+15$

$4 x^{3} - 10 x^{2} + 3 x + 15 = (4 x^{3} - 10 x^{2}) + 6 x + 15$ $4x^3-10x^2+3x+15=(4x^3-10x^2)+6x+15$

$= 2 x^{2} (2 x - 5) + 6 x + 15$ $=2x^2(2x-5)+6x+15$

Now if we take a common factor out of $6 x + 15$ $6x+15$ and get a monomial times $(2 x - 5)$ $(2x-5)$ , then we can finish factoring by grouping. If we get something else, then factoring by grouping won't work.

In this case we get $6 x + 15 = 3 (2 x + 5)$ $6x+15=3(2x+5)$ . Almost!, But close doesn't work in factoring by grouping. So we can't finish this by grouping.

Example 3 You do the test-maker's job.

I want a problem that CAN be factored by grouping.

I start with $12 x^{3} - 28 x^{2}$ $12x^3-28x^2$ So, if it CAN be factored by grouping, the the rest of is has to look like what?

It has to be a monomial times $(3 x - 7)$ $(3x-7)$ .

So finishing with $6 x - 14$ $6x-14$ would work, or $15 x - 35$ $15x-35$ , or I could get tricky and use $- 9 x + 21$ $-9x+21$ . In fact ANY number times $(3 x - 7)$ $(3x-7)$ added to what I already have will give me a polynomial that can be factored by grouping.

$12 x^{3} - 28 x^{2} + k 3 x - k 7$ $12x^3-28x^2+k3x-k7$ for any $k$ $k$ can be factored as:

$12 x^{3} - 28 x^{2} + 3 k x - 7 k = 4 x^{2} (3 x - 7) + k (3 x - 7) = (4 x^{2} + k) (3 x - 7)$ $12x^3-28x^2+3kx-7k=4x^2(3x-7)+k(3x-7)=(4x^2+k)(3x-7)$

Final note: $k = - 1$ $k=-1$ or $k = - 9$ $k=-9$ would make good choices. Because then the fisrt factor is a difference of 2 squares and can be factored.

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Why does factoring polynomials by grouping work?

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