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New Zealand
Level 6 - NCEA Level 1

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}$3x1+x2. 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 $y^5-y^2$y5y2 to get $y^2\left(y^3-1\right)$y2(y31) 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}$3x1+x2, 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}$3x1+x2, the least power is $x^{-2}$x2, since $-2<-1$2<1.

Let's divide $x^{-2}$x2 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)$3x1÷​x2=3(x1÷​x2) and then we can use the division index law $b^m\div b^n=b^{m-n}$bm÷​bn=bmn 

$3\left(x^{-1}\div x^{-2}\right)$3(x1÷​x2) $=$= $3x^{-1-\left(-2\right)}$3x1(2)
  $=$= $3x^{-1+2}$3x1+2
  $=$= $3x^1$3x1
  $=$= $3x$3x

The second term. 

As for dividing our second term $x^{-2}$x2 by $x^{-2}$x2, remember that anything divided by itself equals $1$1.

Therefore, we can factorise our expression by taking out the least power of $x^{-2}$x2 as follows.

$3x^{-1}+x^{-2}$3x1+x2 $=$= $x^{-2}\left(3x+1\right)$x2(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. $x^{-2}$x2 is a smaller number than $x^{-1}$x1, as we can visualise below, and the highest common factor of $x^{-2}$x2 and $x^{-1}$x1 is $x^{-2}$x2.

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}$3x1+x2 $=$= $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 $x^{-2}$x2 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}$3x1+x2 $=$= $\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}$9y818y10 by first factoring out the least power of $y$y.


The least power of $y$y in the terms $9y^{-8}$9y8 and $-18y^{-10}$18y10 is $y^{-10}$y10, since $-10<-8$10<8.

Dividing out $y^{-10}$y10 from the first term $9y^{-8}$9y8 gives:

$9y^{-8}\div y^{-10}$9y8÷​y10 $=$= $9\left(y^{-8}\div y^{-10}\right)$9(y8÷​y10)
  $=$= $9y^{-8-\left(-10\right)}$9y8(10)
  $=$= $9y^{-8+10}$9y8+10
  $=$= $9y^2$9y2

And dividing out $y^{-10}$y10 from the second term $18y^{-10}$18y10 leaves $1$1.

However, there is also a numerical common factor between the two terms $9y^{-8}$9y8 and $18y^{-10}$18y10. They are both divisible by $9$9.

Hence, we divide $9$9 and $y^{-10}$y10 out of both terms, which is the same as dividing $9y^{-10}$9y10 out of both terms. This allows us to factorise the expression as follows.

$9y^{-8}-18y^{-10}$9y818y10 $=$= $9y^{-10}\left(y^2-1\right)$9y10(y21)

Note that you could factorise this answer even further by factorising the difference of two squares $y^2-1$y21.


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 $y-5$y5 from $\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y5)13+(y5)23.

The least power of $y-5$y5 in the expression is $\left(y-5\right)^{-\frac{1}{3}}$(y5)13 and since $-\frac{1}{3}<\frac{2}{3}$13<23.

Dividing out $\left(y-5\right)^{-\frac{1}{3}}$(y5)13 from the first term $\left(y-5\right)^{-\frac{1}{3}}$(y5)13 leaves $1$1.

Dividing out $\left(y-5\right)^{-\frac{1}{3}}$(y5)13 from the second term $\left(y-5\right)^{\frac{2}{3}}$(y5)23 gives:

$\left(y-5\right)^{\frac{2}{3}}\div\left(y-5\right)^{-\frac{1}{3}}$(y5)23÷​(y5)13 $=$= $\left(y-5\right)^{\frac{2}{3}-\left(-\frac{1}{3}\right)}$(y5)23(13)
  $=$= $\left(y-5\right)^{\frac{2}{3}+\frac{1}{3}}$(y5)23+13
  $=$= $\left(y-5\right)^1$(y5)1
  $=$= $y-5$y5

And so we can factorise $\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y5)13+(y5)23 as follows.

$\left(y-5\right)^{-\frac{1}{3}}+\left(y-5\right)^{\frac{2}{3}}$(y5)13+(y5)23 $=$= $\left(y-5\right)^{-\frac{1}{3}}\left(1+\left(y-5\right)\right)$(y5)13(1+(y5))
  $=$= $\left(y-5\right)^{-\frac{1}{3}}\left(y-4\right)$(y5)13(y4)

Further Examples

Question 3

Given the least power of $x$x in the following expression is $-7$7, factorise $x^{-7}-9x^{-5}$x79x5.

Question 4

Factorise $-4p^{-\frac{4}{5}}+12p^{-\frac{9}{5}}$4p45+12p95 by factorising out the lowest power of $p$p.



Generalise the properties of operations with rational numbers, including the properties of exponents


Apply algebraic procedures in solving problems

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