Posted by **Anonymous** on Saturday, March 1, 2014 at 2:16pm.

Can someone show me how to solve these step-by-step? Please? Thank you.

Factor each polynomial by grouping. Check your answer.

1. 2r^2 - 6r + 12 - 4r

2. 14q^2 - 21q + 6 - 4q

3. w^3 -4w^2 + w - 4

4. 2p^3 - 6p^2 + 15 - 5p

- Algebra 1--Step-by-Step -
**Reiny**, Saturday, March 1, 2014 at 2:22pm
1.

2r(r - 3) + 4(3 - r) , ahhh, did you notice that the brackets are opposite? , so ..

= 2r(r-3) - 4(r - 3)

= (r-3)(2r-4)

Use the same "trick" on the others if you have to.

- Algebra 1--Step-by-Step -
**Anonymous**, Saturday, March 1, 2014 at 2:28pm
My textbook says the answer is 2(r-2)(r-3), though.

- Algebra 1--Step-by-Step -
**Reiny**, Saturday, March 1, 2014 at 2:31pm
Surely you could see that 2 was an additional common factor contained in my (2r-4) factor.

- Algebra 1--Step-by-Step -
**Anonymous**, Saturday, March 1, 2014 at 2:35pm
What? I don't understand what you mean.

- Algebra 1--Step-by-Step -
**Reiny**, Saturday, March 1, 2014 at 2:36pm
look at my answer ...

(r-3)(2r-4)

= (r-3)(2)(r-2)

= 2(r-3)(r-2)

- Algebra 1--Step-by-Step -
**Anonymous**, Saturday, March 1, 2014 at 3:09pm
Oh.

My answers to the other equations:

2. ?

3. (w^2 + 1)(w - 4)

4. (2p^2 - 5)(p - 3)

- Algebra 1--Step-by-Step -
**Anonymous**, Saturday, March 1, 2014 at 3:18pm
2. (7q - 2)(2q - 3).

- Algebra 1--Step-by-Step -
**Reiny**, Saturday, March 1, 2014 at 3:20pm
3 and 4 are correct

#2

14q^2 - 21q + 6 - 4q

= 7q(2q - 3) + 2(3-2q)

= 7q(2q-3) - 2(2q-3)

= (2q-3)(7q-2)

- Algebra 1--Step-by-Step -
**Anonymous**, Saturday, March 1, 2014 at 3:24pm
Thank you.

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