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5.01 Simplifying rational expressions


Simplifying by factoring

We've already learned how to simplify fractions that only involve numbers, and algebraic fractions work exactly the same way. Simply find common factors (greatest common factor [GCF] is the fastest way) between the denominator and the numerator and cancel them out until you can't find any more.

For example, $\frac{49st^2}{42t}$49st242t involves numbers and the two variables $s$s and $t$t, so let's look at them all separately. Just looking at the numbers, we see that $7$7 is the $GCF$GCF between $49$49 and $42$42, so we can take it out from both to leave $7$7 and $6$6, respectively. As for $s$s terms, there are no common factors between the denominator and numerator except $1$1 so we leave them. Lastly the $t$t terms have a GCF of $t$t, so we are left with $t$t on the top and $1$1 on the bottom after canceling out. So in the end we should have $\frac{7st}{6}$7st6 after simplifying.


Simplifying by factoring

All of the factoring methods we have learned in the past can greatly help us simplify complicated algebraic fractions. There is, however, something that you must keep in mind. An algebraic fraction will be undefined for any value of the variables that would make the denominator of the fraction equal to zero. So, these values must be restricted from use when you are working with rational expressions.

Did you know?

When we simplify the fraction $\frac{84}{270}$84270 to $\frac{14}{45}$1445, our simplified fraction is fully equivalent. We could go backwards from $\frac{14}{45}$1445 to express the fraction as $\frac{84}{270}$84270 again.

However, when we simplify, say, $\frac{3\left(x-4y\right)}{x-4y}$3(x4y)x4y to $3$3 by canceling a common factor of $x-4y$x4y, our simplified answer is not fully equivalent to what we started with.

This is because in $\frac{3x-12y}{x-4y}$3x12yx4y, the denominator cannot equal zero, so we cannot choose $x$x and $y$y such that $x-4y=0$x4y=0. But when we cancel out the factor $x-4y$x4y, we lose this piece of information.


Worked examples

Question 1

Factor and simplify $\frac{x^2-6x+9}{x-3}$x26x+9x3

Think: about whether the numerator can be factored using quadratic methods or perfect square methods


$x^2-6x+9$x26x+9 can be factored using the perfect square method as $9$9 is a square number, and $6$6 is double $\sqrt{9}=3$9=3

Therefore it becomes $\left(x-3\right)^2$(x3)2

So our fraction can be rewritten as:

$\frac{\left(x-3\right)^2}{x-3}=x-3$(x3)2x3=x3 as $x-3$x3 is a common factor of the numerator and denominator

Reflect: We would have the restriction that $x\ne3$x3 because $x-3$x3 was in the denominator and $x-3\ne0$x30.


Question 2

Factor and simplify $\frac{y+4}{y^2-3y-28}$y+4y23y28

Think about which method to use for the denominator and how the negative $-28$28 will affect it


$y^2-3y-28$y23y28 is a quadratic trinomial but not a perfect square as $-28$28 is not a square number

Its negativity also means the two numbers $a$a and $b$b we need to find in $\left(y+a\right)\left(y+b\right)$(y+a)(y+b) have different signs.

Number pairs that give us $-28$28 are:

$1$1 & $-28$28, $-1$1 & $28$28, $2$2 & $-14$14, $-2$2 & $14$14, $4$4 & $-7$7, $-4$4 & $7$7

The only pair to have a sum of $-3$3 is $4$4 & $-7$7, which must be our $a$a and $b$b

$\left(y+4\right)\left(y-7\right)$(y+4)(y7) must then be the factored form

Our fraction then becomes $\frac{y+4}{\left(y+4\right)\left(y-7\right)}=\frac{1}{y-7}$y+4(y+4)(y7)=1y7 as $\frac{y+4}{y+4}=1$y+4y+4=1. We have the restriction that $y\ne4$y4 from the original denominator.


Practice questions

Question 3

Factor $\frac{6x-16}{12}$6x1612 and simplify.

Question 4

Factor and simplify $\frac{50m^2+70mn}{80m^2}$50m2+70mn80m2.

Question 5

Factor and simplify $\frac{a^2-81}{9-a}$a2819a.


Here are a few more involved examples.

Worked example

Question 6

Simplify $\frac{x^2+5x+4}{x^2-2x-3}$x2+5x+4x22x3.

Think: We factor the numerator and the denominator to obtain $\frac{(x+1)(x+4)}{(x+1)(x-3)}$(x+1)(x+4)(x+1)(x3). The factor $x+1$x+1 is common.

Do: So, we cancel it to see that


This equivalence is correct except when $x=-1$x=1 for which case the original fraction was undefined, so we can say:

$\frac{x^2+5x+4}{x^2-2x-3}=\frac{x+4}{x-3}$x2+5x+4x22x3=x+4x3, where $x\ne-1$x1


Practice questions

Question 7

Simplify $\frac{x^2+xy+xz+yz}{x^2+2xy+y^2}$x2+xy+xz+yzx2+2xy+y2.

Question 8

Factor and simplify $\frac{3a^2+24a+45}{9a^2-81}$3a2+24a+459a281


Further simplifying fractions with polynomial expressions

Remember when factoring algebraic expressions to first try taking out the greatest common factor (GCF) from all terms. Recall that we have many different strategies for factoring polynomial expressions such as difference of squares, perfect squares or grouping.

Worked examples

Question 9

Simplify $\frac{16x^4+40x^3+25x^2}{25-16x^2}$16x4+40x3+25x22516x2.

Think: Firstly, how do we factor the numerator $16x^4+40x^3+25x^2$16x4+40x3+25x2? First we notice, there is a common factor of $x^2$x2, so we are really trying to factor $16x^2+40x+25$16x2+40x+25. We are looking for a binomial factoring $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) for some integers $a$a, $b$b, $c$c, $d$d.

The product of the constant terms $bd$bd will be equal to $25$25. Since $25$25 factors to $5\times5$5×5, we should try $b=5$b=5 and $d=5$d=5 first, which would give $\left(ax+5\right)\left(cx+5\right)$(ax+5)(cx+5).

Now, can we find the $x$x coefficients $a$a and $c$c such that this distributes to $16x^2+40x+25$16x2+40x+25? The product of these will be $16$16, so let's test $a=4$a=4 and $c=4$c=4 first.

$\left(4x+5\right)\left(4x+5\right)=16x^2+20x+20x+25$(4x+5)(4x+5)=16x2+20x+20x+25 so does indeed equal $16x^2+40x+25$16x2+40x+25 when distributed. Hence our factoring is correct. Note that we could have also figured out this factoring by using the quadratic formula to find the zeros of the quadratic.

What about the denominator? Well, notice that $25-16x^2$2516x2 is a difference of two squares so can be factored to $\left(5-4x\right)\left(5+4x\right)$(54x)(5+4x).

Let's now use these factorisations to simplify the fraction.


$\frac{16x^4+40x^3+25x^2}{25-16x^2}$16x4+40x3+25x22516x2 $=$= $\frac{x^2\left(16x^2+40x+25\right)}{25-16x^2}$x2(16x2+40x+25)2516x2 Take out the common factor of $x^2$x2.
  $=$= $\frac{x^2\left(4x+5\right)\left(4x+5\right)}{\left(5-4x\right)\left(5+4x\right)}$x2(4x+5)(4x+5)(54x)(5+4x) Using our factorisations
  $=$= $\frac{x^2\left(4x+5\right)}{5-4x}$x2(4x+5)54x Cancel the common factors $4x+5$4x+5 and $5+4x$5+4x, which are equivalent

Reflect: We have the restriction that $4x+5\ne0$4x+50, so $x\ne\frac{-5}{4}$x54.

Question 10

Simplify $\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k25)410k(2k25)58(2k25)8.

Think: We want to think of$2k^2-5$2k25 as a group.


$\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k25)410k(2k25)58(2k25)8 $=$= $\frac{2\left(2k^2-5\right)^4\left(3-5k\left(2k^2-5\right)\right)}{8\left(2k^2-5\right)^8}$2(2k25)4(35k(2k25))8(2k25)8
  $=$= $\frac{\left(2k^2-5\right)^4\left(3-5k\left(2k^2-5\right)\right)}{4\left(2k^2-5\right)^8}$(2k25)4(35k(2k25))4(2k25)8
  $=$= $\frac{3-5k\left(2k^2-5\right)}{4\left(2k^2-5\right)^4}$35k(2k25)4(2k25)4

Recall our exponent law that states that $b^m\div b^n=b^{m-n}$bm÷​bn=bmn to notice that dividing out $\left(2k^2-5\right)^4$(2k25)4 from $\left(2k^2-5\right)^8$(2k25)8 will leave $\left(2k^2-5\right)^4$(2k25)4.

Finally, we distribute what remains in the numerator to get our final answer of $\frac{3+25k-10k^3}{4\left(2k^2-5\right)^4}$3+25k10k34(2k25)4.




Create and recognize equivalent expressions involving radical and exponential forms of expressions.

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