Factor expressions with negative and rational powers
If you need to, take the time to revise division with negative powers and factorising expressions.
Expressions with Negative Indices
Method One
We can factorise the algebraic expression $3x^3+12x$3x3+12x by taking out the highest common factor from both the terms in the expression. The terms $3x^3$3x3 and $12x$12x both have a highest common factor of $3x$3x, which we can divide out of both the terms. This gives us the factorisation $3x\left(x^2+4\right)$3x(x2+4).
But what if we have an expression with negative indices, such as $3x^{-1}+x^{-2}$3x−1+x−2. How do we factorise this? What is the common factor that we take out of both terms?
Notice that in the previous example we took out the highest common factor, which involved the least power out of $x$x and $x^3$x3, which was $x$x.
We would do the same thing for the expression $y^5-y^2$y5−y2 to get $y^2\left(y^3-1\right)$y2(y3−1) or for the expression $m^{17}+5m^9$m17+5m9 to get $m^9\left(m^8+5\right)$m9(m8+5).
So, in $3x^{-1}+x^{-2}$3x−1+x−2, what is the least power? Can we have a least power between two negative powers?
Well, yes we can. Remember that negative numbers can still be greater or less than other numbers. Numbers further to the left on the number line are always lesser numbers.
Hence, in $3x^{-1}+x^{-2}$3x−1+x−2, the least power is $x^{-2}$x−2, since $-2<-1$−2<−1.
Let's divide $x^{-2}$x−2 out of both terms in the expression.
The first term is $3x^{\left(-1\right)}$3x(−1), so dividing that by $x^{\left(-2\right)}$x(−2)
we get... $3x^{-1}\div x^{-2}=3\left(x^{-1}\div x^{-2}\right)$3x−1÷x−2=3(x−1÷x−2) and then we can use the division index law $b^m\div b^n=b^{m-n}$bm÷bn=bm−n
| $3\left(x^{-1}\div x^{-2}\right)$3(x−1÷x−2) | $=$= | $3x^{-1-\left(-2\right)}$3x−1−(−2) |
| $=$= | $3x^{-1+2}$3x−1+2 | |
| $=$= | $3x^1$3x1 | |
| $=$= | $3x$3x |
The second term.
As for dividing our second term $x^{-2}$x−2 by $x^{-2}$x−2, remember that anything divided by itself equals $1$1.
Therefore, we can factorise our expression by taking out the least power of $x^{-2}$x−2 as follows.
| $3x^{-1}+x^{-2}$3x−1+x−2 | $=$= | $x^{-2}\left(3x+1\right)$x−2(3x+1) |
This may seem strange at first, but it is no different to taking out the highest common factor of two terms in other expressions. $x^{-2}$x−2 is a smaller number than $x^{-1}$x−1, as we can visualise below, and the highest common factor of $x^{-2}$x−2 and $x^{-1}$x−1 is $x^{-2}$x−2.
The highest common factor of any two powers you select above will always be the lesser one, even when negative powers are involved.
Method Two
Notice that the method we used above is equivalent to taking out a fraction as a common factor, because of the reciprocal index law.
| $3x^{-1}+x^{-2}$3x−1+x−2 | $=$= | $3\times\frac{1}{x}+\frac{1}{x^2}$3×1x+1x2 |
This gives us an alternative method for factorising our previous expression. Taking a factor of $x^{-2}$x−2 out of the expression is exactly the same as taking a factor of $\frac{1}{x^2}$1x2 out of the expression.
| $3x^{-1}+x^{-2}$3x−1+x−2 | $=$= | $\frac{3}{x}+\frac{1}{x^2}$3x+1x2 |
| $=$= | $\frac{1}{x^2}\left(3x+1\right)$1x2(3x+1) |
Worked Examples
Question 1
Factorise $9y^{-8}-18y^{-10}$9y−8−18y−10 by first factoring out the least power of $y$y.
Solution
The least power of $y$y in the terms $9y^{-8}$9y−8 and $-18y^{-10}$−18y−10 is $y^{-10}$y−10, since $-10<-8$−10<−8.
Dividing out $y^{-10}$y−10 from the first term $9y^{-8}$9y−8 gives:
| $9y^{-8}\div y^{-10}$9y−8÷y−10 | $=$= | $9\left(y^{-8}\div y^{-10}\right)$9(y−8÷y−10) |
| $=$= | $9y^{-8-\left(-10\right)}$9y−8−(−10) | |
| $=$= | $9y^{-8+10}$9y−8+10 | |
| $=$= | $9y^2$9y2 |
And dividing out $y^{-10}$y−10 from the second term $18y^{-10}$18y−10 leaves $1$1.
However, there is also a numerical common factor between the two terms $9y^{-8}$9y−8 and $18y^{-10}$18y−10. They are both divisible by $9$9.
Hence, we divide $9$9 and $y^{-10}$y−10 out of both terms, which is the same as dividing $9y^{-10}$9y−10 out of both terms. This allows us to factorise the expression as follows.
| $9y^{-8}-18y^{-10}$9y−8−18y−10 | $=$= | $9y^{-10}\left(y^2-1\right)$9y−10(y2−1) |
Note that you could factorise this answer even further by factorising the difference of two squares $y^2-1$y2−1.
EXPRESSIONS WITH Rational INDICES
The approach above extends to rational indices as well. We find the least power and divide it out of both terms.
Worked Examples
Question 3
Factor out the least power of $y-5$y−5 from $\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y−5)−13+(y−5)23.
The least power of $y-5$y−5 in the expression is $\left(y-5\right)^{-\frac{1}{3}}$(y−5)−13 and since $-\frac{1}{3}<\frac{2}{3}$−13<23.
Dividing out $\left(y-5\right)^{-\frac{1}{3}}$(y−5)−13 from the first term $\left(y-5\right)^{-\frac{1}{3}}$(y−5)−13 leaves $1$1.
Dividing out $\left(y-5\right)^{-\frac{1}{3}}$(y−5)−13 from the second term $\left(y-5\right)^{\frac{2}{3}}$(y−5)23 gives:
| $\left(y-5\right)^{\frac{2}{3}}\div\left(y-5\right)^{-\frac{1}{3}}$(y−5)23÷(y−5)−13 | $=$= | $\left(y-5\right)^{\frac{2}{3}-\left(-\frac{1}{3}\right)}$(y−5)23−(−13) |
| $=$= | $\left(y-5\right)^{\frac{2}{3}+\frac{1}{3}}$(y−5)23+13 | |
| $=$= | $\left(y-5\right)^1$(y−5)1 | |
| $=$= | $y-5$y−5 |
And so we can factorise $\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y−5)−13+(y−5)23 as follows.
| $\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y−5)−13+(y−5)23 | $=$= | $\left(y-5\right)^{-\frac{1}{3}}\left(1+\left(y-5\right)\right)$(y−5)−13(1+(y−5)) |
| $=$= | $\left(y-5\right)^{-\frac{1}{3}}\left(y-4\right)$(y−5)−13(y−4) |
Further Examples
Question 3
Given the least power of $x$x in the following expression is $-7$−7, factorise $x^{-7}-9x^{-5}$x−7−9x−5.
Question 4
Factorise $-4p^{-\frac{4}{5}}+12p^{-\frac{9}{5}}$−4p−45+12p−95 by factorising out the lowest power of $p$p.