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.
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.
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(x−4y)x−4y to $3$3 by canceling a common factor of $x-4y$x−4y, our simplified answer is not fully equivalent to what we started with.
This is because in $\frac{3x-12y}{x-4y}$3x−12yx−4y, the denominator cannot equal zero, so we cannot choose $x$x and $y$y such that $x-4y=0$x−4y=0. But when we cancel out the factor $x-4y$x−4y, we lose this piece of information.
Factor and simplify $\frac{x^2-6x+9}{x-3}$x2−6x+9x−3
Think: about whether the numerator can be factored using quadratic methods or perfect square methods
Do:
$x^2-6x+9$x2−6x+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 $\left(x-3\right)^2$(x−3)2
So our fraction can be rewritten as:
$\frac{\left(x-3\right)^2}{x-3}=x-3$(x−3)2x−3=x−3 as $x-3$x−3 is a common factor of the numerator and denominator
Reflect: We would have the restriction that $x\ne3$x≠3 because $x-3$x−3 was in the denominator and $x-3\ne0$x−3≠0.
Factor and simplify $\frac{y+4}{y^2-3y-28}$y+4y2−3y−28
Think about which method to use for the denominator and how the negative $-28$−28 will affect it
Do
$y^2-3y-28$y2−3y−28 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)(y−7) 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)(y−7)=1y−7 as $\frac{y+4}{y+4}=1$y+4y+4=1. We have the restriction that $y\ne4$y≠4 from the original denominator.
Factor $\frac{6x-16}{12}$6x−1612 and simplify.
Factor and simplify $\frac{50m^2+70mn}{80m^2}$50m2+70mn80m2.
Factor and simplify $\frac{a^2-81}{9-a}$a2−819−a.
Here are a few more involved examples.
Simplify $\frac{x^2+5x+4}{x^2-2x-3}$x2+5x+4x2−2x−3.
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)(x−3). The factor $x+1$x+1 is common.
Do: So, we cancel it to see that
$\frac{x^2+5x+4}{x^2-2x-3}=\frac{x+4}{x-3}$x2+5x+4x2−2x−3=x+4x−3
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+4x2−2x−3=x+4x−3, where $x\ne-1$x≠−1
Simplify $\frac{x^2+xy+xz+yz}{x^2+2xy+y^2}$x2+xy+xz+yzx2+2xy+y2.
Factor and simplify $\frac{3a^2+24a+45}{9a^2-81}$3a2+24a+459a2−81
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.
Simplify $\frac{16x^4+40x^3+25x^2}{25-16x^2}$16x4+40x3+25x225−16x2.
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$25−16x2 is a difference of two squares so can be factored to $\left(5-4x\right)\left(5+4x\right)$(5−4x)(5+4x).
Let's now use these factorisations to simplify the fraction.
Do:
$\frac{16x^4+40x^3+25x^2}{25-16x^2}$16x4+40x3+25x225−16x2 | $=$= | $\frac{x^2\left(16x^2+40x+25\right)}{25-16x^2}$x2(16x2+40x+25)25−16x2 | 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)(5−4x)(5+4x) | Using our factorisations | |
$=$= | $\frac{x^2\left(4x+5\right)}{5-4x}$x2(4x+5)5−4x | 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+5≠0, so $x\ne\frac{-5}{4}$x≠−54.
Simplify $\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k2−5)4−10k(2k2−5)58(2k2−5)8.
Think: We want to think of$2k^2-5$2k2−5 as a group.
Do:
$\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k2−5)4−10k(2k2−5)58(2k2−5)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(2k2−5)4(3−5k(2k2−5))8(2k2−5)8 |
$=$= | $\frac{\left(2k^2-5\right)^4\left(3-5k\left(2k^2-5\right)\right)}{4\left(2k^2-5\right)^8}$(2k2−5)4(3−5k(2k2−5))4(2k2−5)8 | |
$=$= | $\frac{3-5k\left(2k^2-5\right)}{4\left(2k^2-5\right)^4}$3−5k(2k2−5)4(2k2−5)4 |
Recall our exponent law that states that $b^m\div b^n=b^{m-n}$bm÷bn=bm−n to notice that dividing out $\left(2k^2-5\right)^4$(2k2−5)4 from $\left(2k^2-5\right)^8$(2k2−5)8 will leave $\left(2k^2-5\right)^4$(2k2−5)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+25k−10k34(2k2−5)4.
Create and recognize equivalent expressions involving radical and exponential forms of expressions.