Core practice ← Back to lesson

Multiplying and Dividing Rational Expressions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 Two quadratic denominators

Write $\frac{3x+2}{x^2+5}\cdot\frac{5}{x^2+7}$ as one fraction in lowest terms for real $x$ , retaining its domain.
1. Hint 1
  Multiplication combines numerators and denominators directly.
2. Hint 2
  Check whether either denominator can be zero for a real input, and compare the numerator's factors with the denominators'.
Answer

$\frac{5(3x+2)}{(x^2+5)(x^2+7)}$ ; all real $x$ .

Full solution

Both denominators are positive for every real input.
Multiplying straight across gives
$\frac{5(3x+2)}{(x^2+5)(x^2+7)}$

The only zero of $3x+2$ is $-\frac23$ , where neither denominator is zero, so $3x+2$ shares no factor with them.
Neither quadratic has a real zero, so neither factors further over the reals, and the constant $5$ has nothing to cancel against.
The domain is all real numbers.
No common denominator had to be created before multiplication.

Answer

$\frac{5(3x+2)}{(x^2+5)(x^2+7)}$ ; all real $x$ .

Key idea

Multiplying rational expressions requires no common-denominator step.
Problem 1: Mark complete
Problem 2 A missing factor

What rational factor must multiply $(x-1)^2$ to produce $\frac{x-1}{x^2+1}$ for every real $x\ne1$ ? Give it in lowest terms.
1. Hint 1
  Undo multiplication by a nonzero quantity with its reciprocal.
2. Hint 2
  Count how many copies of $x-1$ remain after removing a common factor.
Answer

$\frac{1}{(x^2+1)(x-1)}$ , with $x\ne1$ .

Full solution

On the stated domain, $(x-1)^2$ is nonzero.
The needed factor is the desired product divided by that square.
This gives
$\frac{x-1}{(x^2+1)(x-1)^2}$

Cancel one nonzero copy of $x-1$ to obtain $1/((x^2+1)(x-1))$ .
Multiplying this by $(x-1)^2$ returns $(x-1)/(x^2+1)$ at every required input.
The factor is defined exactly for real $x\ne1$ .

Answer

$\frac{1}{(x^2+1)(x-1)}$ , with $x\ne1$ .

Key idea

Dividing by a polynomial requires excluding its zeros even when it has no printed denominator.
Problem 2: Mark complete
Problem 3 A zero dividend

Simplify $\frac{0}{x-2}\div\frac{x+1}{x+4}$ for real $x$ , giving its exact domain.
1. Hint 1
  A zero dividend does not remove the conditions for division to be defined.
2. Hint 2
  Check both printed denominators and the divisor's numerator.
Answer

$0$ , with $x\ne2,-4,-1$ .

Full solution

The dividend requires $x\ne2$ .
The divisor requires $x\ne-4$ to be defined and $x\ne-1$ to be nonzero.
On that domain the quotient is
$0\cdot\frac{x+4}{x+1}=0$

At each of the three excluded values an original division is invalid, even though the reduced constant $0$ has no visible restriction.

Answer

$0$ , with $x\ne2,-4,-1$ .

Key idea

A zero dividend still requires a defined, nonzero divisor.
Problem 3: Mark complete
Problem 4 Area of a panel

A rectangular panel has width $\frac{x^2+x-56}{x+8}$ cm and length $\frac{x^2+7x-8}{x-7}$ cm, with $x>7$ . Find its area in simplified form, then evaluate it at $x=9$ .
1. Hint 1
  Area multiplies the two measurements.
2. Hint 2
  Factor both quadratic numerators so the cancellations appear before multiplication.
Answer

$(x+8)(x-1)$ square cm; $136$ square cm when $x=9$ .

Full solution

The condition $x>7$ makes both denominators nonzero and both measurements positive.
Factoring, the numerators are $(x+8)(x-7)$ and $(x+8)(x-1)$ , so the area is
$\frac{(x+8)(x-7)}{x+8}\cdot\frac{(x+8)(x-1)}{x-7}$
The factors $x+8$ and $x-7$ cancel, giving
$A=(x+8)(x-1)$

At $x=9$ the width is $\frac{34}{17}=2$ cm and the length is $\frac{136}{2}=68$ cm, so their product is $136$ square cm, agreeing with $(9+8)(9-1)$ .

Answer

$(x+8)(x-1)$ square cm; $136$ square cm when $x=9$ .

Key idea

Factoring polynomial measurements can simplify their product before any numerical evaluation.
Problem 4: Mark complete
Problem 5 Comparing two ratios

For real $x$ , let $A(x)=\frac{x^2+x+3}{x^2+x-30}$ and $B(x)=\frac{x^2+x+3}{x+6}$ . Simplify $A(x)\div B(x)$ and identify which original restriction is no longer visible in the final denominator.
1. Hint 1
  The divisor's numerator has no real zero, but both original denominators still matter.
2. Hint 2
  Factor $x^2+x-30$ and multiply by the reciprocal of $B$ .
Answer

$\frac1{x-5}$ , with $x\ne-6,5$ ; the hidden restriction is $x\ne-6$ .

Full solution

The original denominators are $(x+6)(x-5)$ and $x+6$ , which exclude $-6$ and $5$ .
The divisor's numerator is $x^2+x+3=(x+\frac12)^2+\frac{11}4$ , which is positive, so it adds no further excluded real input.
On this domain,
$\begin{aligned}&\frac{x^2+x+3}{(x+6)(x-5)}\cdot\frac{x+6}{x^2+x+3}\\&=\frac1{x-5}\end{aligned}$

The denominator $x-5$ still excludes $5$ .
The restriction at $-6$ survives only in writing; the reduced expression alone would give $-\frac1{11}$ there.

Answer

$\frac1{x-5}$ , with $x\ne-6,5$ ; the hidden restriction is $x\ne-6$ .

Key idea

A quotient can hide an original restriction even when its divisor never equals zero on its own domain.
Problem 5: Mark complete
Problem 6 Just enough copies

For a positive integer $n$ , consider $\frac{(x-6)^3}{(x+7)^2}\cdot\frac{(x+7)^n}{x-6}$ over the real numbers. Find the smallest $n$ for which the product simplifies to a polynomial, give that polynomial, and state the domain of the original product.
1. Hint 1
  Count the copies of each factor above and below the bar after multiplying.
2. Hint 2
  Every copy of $x+7$ in the denominator must be canceled for the result to be a polynomial, but canceling never removes an original restriction.
Answer

$n=2$ ; the product simplifies to $(x-6)^2$ , with $x\ne-7,6$ .

Full solution

The original denominators exclude $-7$ and $6$ , whatever $n$ is.
Multiplying gives
$\frac{(x-6)^3(x+7)^n}{(x+7)^2(x-6)}$

On the domain, one copy of $x-6$ cancels, leaving $(x-6)^2$ above.
For $n=1$ one copy of $x+7$ remains below, giving $\frac{(x-6)^2}{x+7}$ , which is not a polynomial.
For $n=2$ both copies cancel, giving $(x-6)^2$ .

So the smallest $n$ is $2$ , and the product equals $(x-6)^2$ for $x\ne-7,6$ .
The polynomial alone is defined everywhere, so the two restrictions must be stated beside it.

Answer

$n=2$ ; the product simplifies to $(x-6)^2$ , with $x\ne-7,6$ .

Key idea

A product can simplify to a polynomial while its original denominators still exclude inputs.
Problem 6: Mark complete
Problem 7 Three linked expressions

Simplify $\frac{3x^2+13x-10}{2x^2+13x+20}\cdot\frac{2x+5}{3x-2}\div(x+5)$ for real $x$ , retaining every restriction.
1. Hint 1
  Treat the final division as multiplication by a reciprocal, after recording its restriction.
2. Hint 2
  Factor both quadratics, then compare their factors with the second fraction and the final divisor.
Answer

$\frac1{x+4}$ , with $x\ne-5,-4,-\frac52,\frac23$ .

Full solution

Factoring, $3x^2+13x-10=(3x-2)(x+5)$ and
$2x^2+13x+20=(2x+5)(x+4).$
The denominators exclude $-\frac52$ , $-4$ and $\frac23$ , and the divisor $x+5$ must be nonzero, which excludes $-5$ .

Writing the division as multiplication by $\frac1{x+5}$ gives
$\frac{(3x-2)(x+5)(2x+5)}{(2x+5)(x+4)(3x-2)(x+5)}$
On the domain the factors $3x-2$ , $x+5$ and $2x+5$ are nonzero and cancel, leaving $\frac1{x+4}$ with all four restrictions retained.
At $x=0$ the original is $\frac{-10}{20}\cdot\frac5{-2}\div5=\frac14$ , matching $\frac1{0+4}$ .

Answer

$\frac1{x+4}$ , with $x\ne-5,-4,-\frac52,\frac23$ .

Key idea

A sequence of products and quotients carries every original restriction to its final form.
Problem 7: Mark complete
Problem 8 Where the parentheses go

Let $P=\frac3{x-8}$ , $Q=\frac{x+6}{x-8}$ and $R=\frac{x+6}{x-9}$ . A student says $(P\div Q)\div R$ and $P\div(Q\div R)$ are the same expression. Simplify both, give each domain, and decide whether the student is right.
1. Hint 1
  Work inside the parentheses first, and record every restriction each division adds.
2. Hint 2
  A divisor must be defined and nonzero; for $Q\div R$ as a divisor, that includes where $Q\div R$ itself is zero.
3. Hint 3
  Compare the two simplified forms at one allowed input.
Answer

$(P\div Q)\div R=\frac{3(x-9)}{(x+6)^2}$ and $P\div(Q\div R)=\frac3{x-9}$ , each with $x\ne-6,8,9$ . They are not the same expression.

Full solution

Every part must be defined, so $x\ne8$ and $x\ne9$ .
The divisors $Q$ and $R$ must be nonzero, so $x\ne-6$ .

First expression: $P\div Q=\frac3{x-8}\cdot\frac{x-8}{x+6}=\frac3{x+6}$ .
Then
$\frac3{x+6}\cdot\frac{x-9}{x+6}=\frac{3(x-9)}{(x+6)^2}$
with
$x\ne-6,8,9.$

Second expression: $Q\div R=\frac{x+6}{x-8}\cdot\frac{x-9}{x+6}=\frac{x-9}{x-8}$ , which is zero only at the already excluded $9$ .
Then
$\frac3{x-8}\cdot\frac{x-8}{x-9}=\frac3{x-9}$
with
$x\ne-6,8,9.$

The domains agree, but the values do not: at $x=0$ the first is $\frac{3(-9)}{36}=-\frac34$ and the second is
$\frac3{-9}=-\frac13.$
So the two expressions are not the same.

Answer

$(P\div Q)\div R=\frac{3(x-9)}{(x+6)^2}$ and $P\div(Q\div R)=\frac3{x-9}$ , each with $x\ne-6,8,9$ . They are not the same expression.

Key idea

Division is not associative, so the grouping of a chain of quotients can change its value.
Problem 8: Mark complete
Problem 9 Two equivalent operations

For real $x\ne0$ , a student says that $\frac{1}{x^2+1}\div x$ and $\frac{1}{x}\cdot\frac{1}{x^2+1}$ have the same value and the same domain. Is the statement true? Explain.
1. Hint 1
  Write the division using the reciprocal of the polynomial divisor.
2. Hint 2
  Compare every original denominator and the divisor condition.
Answer

True; both equal $\frac1{x(x^2+1)}$ and have domain $x\ne0$ .

Full solution

The first expression requires $x\ne0$ because the divisor is $x$ , while $x^2+1$ is never zero for real inputs.
The second requires the same condition from its denominator $x$ .

On this common domain,
$\frac1{x^2+1}\div x=\frac1{x(x^2+1)}$
Multiplying the second pair gives the same fraction.
Thus both the value and domain agree.

Answer

True; both equal $\frac1{x(x^2+1)}$ and have domain $x\ne0$ .

Key idea

A division and its reciprocal product agree when all their original restrictions agree.
Problem 9: Mark complete
Problem 10 One order defined, the other not

Construct rational expressions $A$ and $B$ , each with denominator $x^2+11$ , so that $A\div B$ is defined for every real $x$ but $B\div A$ is undefined at exactly $x=-10$ and $x=9$ . Give both quotients in simplest form with their domains, and explain why your example works.
1. Hint 1
  A quotient needs its divisor defined and nonzero; with this denominator, every expression is defined for all real $x$ .
2. Hint 2
  So the question is where each expression is zero. Which one must never be zero, and where must the other be zero?
Answer

One example: $A=\frac{(x+10)(x-9)}{x^2+11}$ and $B=\frac1{x^2+11}$ . Then $A\div B=(x+10)(x-9)$ for all real $x$ , and $B\div A=\frac1{(x+10)(x-9)}$ with $x\ne-10,9$ . Other valid examples are accepted.

Full solution

The denominator $x^2+11$ is positive, so $A$ and $B$ are both defined for every real $x$ .
A quotient then fails exactly where its divisor is zero.

For $A\div B$ to be defined everywhere, $B$ must never be zero; $B=\frac1{x^2+11}$ works.
For $B\div A$ to fail exactly at $-10$ and $9$ , $A$ must be zero exactly there; $A=\frac{(x+10)(x-9)}{x^2+11}$ works.

Then
$\begin{aligned}&A\div B\\&=\frac{(x+10)(x-9)}{x^2+11}\cdot\frac{x^2+11}1\\&=(x+10)(x-9)\end{aligned}$
for every real $x$ .
And
$\begin{aligned}&B\div A\\&=\frac1{x^2+11}\cdot\frac{x^2+11}{(x+10)(x-9)}\\&=\frac1{(x+10)(x-9)}\end{aligned}$
with $x\ne-10,9$ .

Answer

One example: $A=\frac{(x+10)(x-9)}{x^2+11}$ and $B=\frac1{x^2+11}$ . Then $A\div B=(x+10)(x-9)$ for all real $x$ , and $B\div A=\frac1{(x+10)(x-9)}$ with $x\ne-10,9$ . Other valid examples are accepted.

Key idea

Reversing a quotient moves the nonzero condition onto the other expression, so the two orders can have different domains.
Problem 10: Mark complete

← Back to lesson