Simplifying Rational Expressions: Core practice

Factor both parts.

The numerator has a common factor of $3$:

The denominator has a common factor of $6$, leaving $x^2+5x-266$, and the numbers $-14$ and $19$ multiply to $-266$ and add to $5$, so the denominator is $6(x-14)(x+19)$.

Read the excluded values off that original denominator now.

It is zero at $14$ and at $-19$.

The factors $14-x$ and $x-14$ are the same two quantities subtracted in opposite order, so $14-x=-(x-14)$, and the numerator is $-3(x-14)$.

On the domain $x-14$ is nonzero, so it divides out, and the numerical factors reduce the way $\frac{3}{6}$ reduces to $\frac12$:

The restriction $x\ne-19$ is still visible, because $x+19$ is still downstairs, but $x\ne14$ is not, so it is written beside the answer.

A check at $x=1$: the original is $\frac{39}{-1560}$, which is $-\frac{1}{40}$, and the simplified form is $-\frac{1}{2\cdot20}$, also $-\frac{1}{40}$.

The numerator is $12(x^2+1)$ and the denominator is $18(x^2+6)$.

Reducing $12/18$ gives

The denominator is positive for all real $x$, so nothing is excluded.

Neither $x^2+1$ nor $x^2+6$ has a real zero, so neither has a linear factor with real coefficients, and a common non-constant factor with real coefficients would have to be a quadratic dividing both.

A quadratic that divides another quadratic is a constant multiple of it, so such a factor would be a constant multiple of each, and since both are monic that would force $x^2+1$ and $x^2+6$ to be equal, which they are not.

Nothing further cancels.

The quotient is $(4x^2+11x-3)/(x+3)$.

Its denominator is positive for every $x>1$, so nothing on the model's range is excluded.

The numerator factors as $(4x-1)(x+3)$, and dividing out the common factor leaves

for $x>1$.

At $x=5$ the rate is $19$ grams per cm.

The original quantities are $152$ grams and $8$ cm, whose quotient is also $19$.

The only monic quadratic whose zeros are exactly $-16$ and $13$ is $(x+16)(x-13)$.

Multiplying it by the desired output gives the numerator, and on the permitted inputs the quotient is that output:

The original denominator is zero precisely at $-16$ and $13$, so these exclusions are exact, and no other input is excluded.

Expand the two cubes.

The cubic terms cancel, the linear terms cancel, and the remaining numerator is

The denominator is positive for every real $x$.

The expression is therefore $1$ throughout the real line.

At $x=0$, the numerator and denominator are both $2$, providing a quick check.

The denominator is positive for every real $x$, so the domain is the full real line.

Let $N$ denote the numerator.

Division gives quotient $x^2+a-1$ and remainder $2-a$, since

Complete cancellation requires the remainder to be zero, so $a=2$.

The numerator then becomes $(x^2+1)^2$, and the reduced expression is $x^2+1$.

There are no excluded real inputs.

The original denominator $x+17$ is zero only at $-17$, so that is the one excluded value.

The numerator has a common factor of $4$:

On the domain $x+17$ is nonzero, so it divides out of the whole numerator and the whole denominator:

So the reported $4$ is right, provided it carries $x\ne-17$.

The method is another matter.

Applied to $\frac{4x+20}{x+17}$, crossing out $20$ and $17$ and then the $x$s again reports $4$.

But $x=1$ is allowed there, and the value at $1$ is $\frac{24}{18}=\frac43$, not $4$.

A method that gives a wrong result is not valid, so the right result above came from the factor $4(x+17)$, not from the crossing-out.

Crossing out is not division.

Dividing the whole numerator $4x+68$ by $17$ gives $\frac{4x}{17}+4$, not $4x$: the $68$ does not disappear, it becomes $4$.

Deleting addends is not a valid rule, even when it lands on a right answer.

Dividing the whole numerator and the whole denominator by the same common factor keeps the value on the original domain, and every input the original excluded stays excluded.

Read each domain off its original denominator.

The denominator of $A$ is $(2x-5)(x-4)$, which is zero at $\frac52$ and at $4$.

The denominator of $B$ is $(3x-7)(x-4)$, which is zero at $\frac73$ and at $4$.

So both are undefined at $4$.

At $\frac52$ only $A$ is undefined, so $B$ is defined there; at $\frac73$ only $B$ is undefined, so $A$ is defined there.

Those two inputs are the only ones at which exactly one of them is defined.

A check at $x=\frac52$: the denominator of $B$ is $\frac12\cdot\left(-\frac32\right)$, which is not zero, so $B$ is defined there, and its value is that of $\frac{x+18}{x-4}$, which is $-\frac{41}{3}$.

At every input other than $4$, $\frac52$ and $\frac73$, both are defined and both equal $\frac{x+18}{x-4}$.

Simplifying preserved their values, not their domains, and the domains are what tell them apart.

For the example, the denominator $(x-15)(x+21)$ is zero at $15$ and at $-21$, and nowhere else.

At every allowed input the factor $(x-15)(x+21)$ is nonzero and divides out:

The reduced constant $5$ is defined at $15$ and at $-21$, but the original is $\frac{0}{0}$ at both and is undefined there.

So a constant reduced value does not determine the original domain, and the conclusion is not forced.

For the general claim, let $F$ be everything the numerator and denominator have in common.

Canceling it leaves only constants, so the numerator is $aF$ and the denominator is $bF$ for nonzero constants $a$ and $b$, and the reduced value is $\frac{a}{b}$.

At an excluded input the denominator $bF$ is zero, and $b$ is not, so $F$ is zero there.

Then the numerator $aF$ is zero too.