**Index laws**

Practice questions, and answers, here

**Common factors**

Watch this video for the basics: video on taking out common factors. Do these worksheets to get good at taking out common factors in algebra without fractions.

When we’re doing series or induction problems, we often have to take out common factors involving fractions.

Example: ^{1}⁄_{6}n(n+1)(2n+1) + ^{1}⁄_{2}n(n+1)

Choose the common factor you take out so that all the fraction bit is now in the common factor, and in the big bracket there are only whole numbers.

So take ^{1}⁄_{6}n(n+1) as the common factor.

^{1}⁄_{6}n(n+1)(2n+1) + ^{1}⁄_{2}n(n+1) = ^{1}⁄_{6}n(n+1) [(2n+1)+3] … because ^{1}⁄_{6}n(n+1) divides into ^{1}⁄_{2}n(n+1) three times, same as ^{1}⁄_{6} of anything divides into ^{1}⁄_{2} of that same thing three times.

So

^{1}⁄_{6}n(n+1)(2n+1) + ^{1}⁄_{2}n(n+1) = ^{1}⁄_{6}n(n+1) [(2n+1)+3]

= ^{1}⁄_{6}n(n+1) (2n+4)

**Practice**

Take out common factors and simplify these:

1. ^{1}⁄_{6}n(n+1)(2n+1) + n(n+1)

2. ^{1}⁄_{6}n(n+1)(2n+1) + 2n(n+1)

3. ^{1}⁄_{4}n^{2}(n+1)^{2} + ^{1}⁄_{6}n(n+1)(2n+1) + n(n+1)

4. ^{2}⁄_{4}n^{2}(n+1)^{2} + ^{1}⁄_{6}n(n+1)(2n+1)

5. ^{1}⁄_{4}n^{2}(n+1)^{2} + ^{2}⁄_{6}n(n+1)(2n+1)

6. ^{1}⁄_{4}n^{2}(n+1)^{2} + ^{3}⁄_{6}n(n+1)(2n+1) + n(n+1)

7. ^{1}⁄_{6}n(n+1)(2n+1) + n

8. ^{1}⁄_{6}n(n+1)(2n+1) − n

9. ^{1}⁄_{6}n(n+1)(2n+1) − 2n

10. ^{1}⁄_{6}n(n+1)(2n+1) + ^{1}⁄_{2}n(n+1) − n

11. ^{1}⁄_{2}(k+1)(k+2) − ^{1}⁄_{2}k(k+1)

12. ^{1}⁄_{6}(k+1)(k+2)(2k+3) − ^{1}⁄_{6}k(k+1)(2k+1)

13. ^{1}⁄_{4}(k+1)^{2}(k+2)^{2} − ^{1}⁄_{4}k^{2}(k+1)^{2}

More practice: