# Algebra 1--Step-by-Step

posted by on .

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

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 - ,

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 - ,

My textbook says the answer is 2(r-2)(r-3), though.

• Algebra 1--Step-by-Step - ,

Surely you could see that 2 was an additional common factor contained in my (2r-4) factor.

• Algebra 1--Step-by-Step - ,

What? I don't understand what you mean.

• Algebra 1--Step-by-Step - ,

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

• Algebra 1--Step-by-Step - ,

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 - ,

2. (7q - 2)(2q - 3).

• Algebra 1--Step-by-Step - ,

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 - ,

Thank you.