Simplifying Rational Expressions

Simplifying Rational Expressions exercise

• Explain how to simplify the following expression.

  Hints

  You can only cancel out common factors of the numerator and the denominator. For example,

  $\frac{(2)(3)(x)(y)}{(3)(y)(5)}=\frac{(2)(\not 3)(x)(\not y)}{(\not 3)(\not y)(5)}=\frac{2x}{5}$.

  You can cancel out the greatest common factor:

  $\frac{(3)(x)(y)}{(3)(x)}=\frac{(\not 3)(x)(y)}{(\not 3)(x)}=\frac{(x)(y)}{(x)}=\frac{(\not x)(y)}{(\not x)}=y$.

  Solution

  To simplify rational expressions we first have to look for the greatest common factor of the numerator and the denominator.

  First we factor the numerator $32x^2=(8)(4)(x)(x)$ and the denominator $24x=(8)(3)(x)$, and we can then see that the greatest common factor is $(8)(x)$.

  We can then cancel out the greatest common factor, to get:

  $\frac{32x^2}{24x}=\frac{(8)(4)(x)(x)}{(8)(3)(x)}=\frac{(\not 8)(4)(\not x)(x)}{(\not 8)(3)(\not x)}=\frac{4x}{3}$.

• Simplify the rational expression.

  Hints

  You can check a factorization by using the distributive property.

  Let's have a look at the following example: $9x+12x^2=(3x)(3+4x)$.

  The other way around we get

  $(3x)(3+4x)=(3x)(3)+(3x)(4x)=9x+12x^2$ $~~~~~$✓

  The FOIL method is a way of remembering the order of multiplying two binomials:

  $(x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12$.

  To factor a quadratic expression, $x^2+bx+c$, look for the factors of $c$ which sum to $b$.

  Solution

  First we look at the numerator; we factor out $6x^2$ to get $6x^3+42x^2=(6x^2)(x+7)$.

  Next we factor the denominator:

  • Factoring out $2$ we get $2x^2+26x+84=(2)(x^2+13x+42)$.
  • We look for the factors of $42$ which sum to $13$, which are $6$ and $7$. Thus we have that $x^2+13x+42=(x+7)(x+6)$.

  We can then see that the greatest common factor of the numerator and the denominator is $(2)(x+7)$, which we can cancel out to get

  $\frac{6x^3+42x^2}{2x^2+26x+84}=\frac{(6x^2)(x+7)}{(2)(x+7)(x+6)}=\frac{3x^2}{x+6}$.

• Decide the terms to be cancelled out.

  Hints

  Never cancel out any number added or subtracted to a variable term.

  You can only cancel out factors which appear in both the numerator and the denominator.

  Let's have a look at the example beside: the factors $2$ as well as $a$ are common factors of the numerator and denominator, so they can be cancelled out.

  Solution

  Remember, you cannot cancel out any numbers added or subtracted to a variable term. You can only cancel out entire polynomial factor.

  1. $\frac{(3)(4)(x)(z)}{(4)(x)(y)}=\frac{(3)(\not 4)(\not x)(z)}{(\not 4)(\not x)(y)}=\frac{3z}y$
  2. $\frac{(x+3)(x+1)(z)}{(3)(x+1)(z)}=\frac{(x+3)(x \not +1)(\not z)}{(3)(x \not +1)(\not z)}=\frac{x+3}3$. One cannot simplify this expression any further, as $3$ appears as a summand in the numerator.
  3. $\frac{(x+5)(x+5)(x+2)}{(5)(6)(x+5)}$. Here, one can only cancel out the binomial $x+5$ to get the resulting rational expression, $\frac{(x+5)(x+2)}{30}$.

• Determine the simplified rational expression.

  Hints

  You can factor a difference of squares in the way you see beside.

  To factor a trinomial, $x^2+bx+c$, find the factors of $c$ which sum to $b$.

  Using the distributive property, the resulting rational expression is given by $\frac{7x-14}{3x+3}$.

  Solution

  First we look at the numerator and factor out $42$ to get $42x^2-168=(42)(x^2-4)$. The binomial inside the parentheses is the difference of two squares, $x^2-2^2$. This can be written as the product of two binomials, $x^2-2^2=(x+2)(x-2)$.

  Next we investigate the denominator. Factoring out $18$ gives us

  $18x^2+54x+36=(18)(x^2+3x+2)$.

  To factor the trinomial we use reverse FOIL; we look for the factors of $2$ which sum to $3$, namely $1$ and $2$. This gives us

  $x^2+3x+2=(x+1)(x+2)$.

  Looking at the numbers $42$ in the numerator and $18$ in the denominator, we recognize the common factor $6$. Thus we rewrite the expression above as follows:

  $\frac{42x^2-168}{18x^2+54x+36}=\frac{(6)(7)(x+2)(x-2)}{(3)(6)(x+1)(x+2)}$.

  Now we can realize the greatest common factor, $(6)(x+2)$, which we cancel out to get

  $\frac{(6)(7)(x+2)(x-2)}{(3)(6)(x+1)(x+2)}=\frac{(7)(x-2)}{(3)(x+1)}$.

  We still use the distributive property to get

  $\frac{(7)(x-2)}{(3)(x+1)}=\frac{7x-14}{3x+3}$.

  This expression can't be simplified further, so we are done.

• Describe what can be done to simplify rational expressions.

  Hints

  For example, $\frac{3+5}5=\frac85$.

  Using FOIL multiplication, we have that $(x-5)(x-5)=x^2-5x-5x+25=x^2-10x+25$.

  If you divide $2$ by $2$ you get $1$.

  Solution

  Another way of simplifying expressions is to look for matching binomial pairs.

  Let's have a look at the following expression: $\frac{x-5}{x^2-10x+25}$.

  The numerator is already simplified. The denominator can be factored: $x^2-10x+25=(x-5)(x-5)$.

  One can't cancel out $-5$; you can only cancel out entire polynomial factors.

  We can cancel out $x-5$ to get $\frac{x-5}{x^2-10x+25}=\frac{x-5}{(x-5)(x-5)}=\frac1{x-5}$.

  Keep in mind that any number divided by itself is equal to $1$.

• Examine if the given rational expressions can be simplified.

  Hints

  Factoring the difference of two squares in general looks like $a^2-b^2=(a+b)(a-b)$.

  You can only cancel out entire polynomials as a factor.

  You can factor a quadratic expression $x^2+bx+c$ by looking for the pairs of factors of $c$ which sum to $b$.

  Solution

  Let's start with $\frac{6x^2+4x}{3x^2+5x+2}$:

  • Factor $2x$ out of the numerator to get $6x^2+4x=(2x)(3x+2)$.
  • Use reverse FOIL to factor the denominator, $3x^2+5x+2=(3x+2)(x+1)$.

  We can then see that $3x+2$ is a common factor of the numerator and denominator, and so we cancel it out:

  $\frac{6x^2+4x}{3x^2+5x+2}=\frac{(2x)(3x+2)}{(3x+2)(x+1)}=\frac{2x}{x+1}$.

  Next we examine $\frac{3x^3+3x^2}{x^3+2x^2+x}$:

  • $3x^3+3x^2=(3x^2)(x+1)=(3x)(x)(x+1)$.
  • $x^3+2x^2+x=(x)(x^2+2x+1)=(x)(x+1)(x+1)$.

  Thus we can see that $(x)(x+1)$ is a common factor of the numerator and denominator, and so we cancel it out:

  $\frac{3x^3+3x^2}{x^3+2x^2+x}=\frac{(3x)(x)(x+1)}{(x)(x+1)(x+1)}=\frac{3x}{x+1}$.

  Now we have $\frac{4x^2}{2x^2+4x}$;

  • $4x^2=(2x)(2x)$.
  • $2x^2+4x=(2x)(x+2)$.

  This gives us the greatest common factor $2x$, which we cancel out:

  $\frac{4x^2}{2x^2+4x}=\frac{(2x)(2x)}{(2x)(x+2)}=\frac{2x}{x+2}$.

  The last expression is $\frac{4x^2-8x}{x^2-4}$:

  • $4x^2-8x=(4x)(x-2)$.
  • $x^2-4=x^2-2^2=(x+2)(x-2)$.

  So we get $\frac{4x^2-8x}{x^2-4}=\frac{(4x)(x-2)}{(x+2)(x-2)}=\frac{4x}{x+2}$ by cancelling out $x-2$.