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Question

1. Find the error in the work shown below. Explain the error completely and what should have been done instead. Finally, enter the correct simplification process.

2. Divide the expression shown, then simplify completely.



Answer 1

When you're simplifying a rational expression (polynomial fraction), anything you cancel out needs to be multiplying the entire numerator and entire denominator. So for example,

(a x b)/(a x c) = b/c

but

(a + b)/(a + c) ≠ b/c

So if a term is multiplying the entire numerator and denominator, it can be canceled, but if the numerator is a sum of terms and it only shows up in one term of the numerator, it can't be canceled out.

This means that simplifying rational expressions or dividing polynomials involves factoring the polynomial. Let's factor the numerator in the problem you are supposed to correct.

x² + 2x - 8

We are looking for two numbers that multiply to equal negative 8, but add to equal positive 2. Since they have to multiply to equal a negative number, one of them must be positive and one must be negative. -4 and 2 will work. You can "FOIL" out this expression to check that it works.

(x + 4)(x - 2)

Now the denominator. We are looking for two numbers that multiply to -16 and add to 6. 8 and -2 will work.

x² + 6x - 16 = (x + 8)(x - 2)

So the whole rational expression is,

[(x + 4)(x - 2)]/[(x + 8)(x - 2)]

(x - 2) is multiplying everything in the numerator and everything in the denominator. We can cancel it out. The correct solution is

(x + 4)/(x + 8)

Same approach to the second problem, for the most part. The other info we need to remember for the second problem is that dividing by a fraction is the same as multiplying by the reciprocal (flipped) fraction. So,

(x² - 9)/(x² + x - 2) ÷ (x² + x - 12)/(x² + 6x + 8) = (x² - 9)/(x² + x - 2) × (x² + 6x + 8)/(x² + x - 12)

Now for the factoring. When factoring x² - 9, it might be helpful to write it as x² +0x - 9, since you are probably used to factor quadratic expressions with 3 terms. So we want two numbers multiplying to -9 and adding to 0. That would be -3 and 3. So the first numerator is (x + 3)(x - 3).

Now the first denominator, x² + x - 2. We want two numbers that will multiply to -2 and add to 1. So 2 and -1. This gives us (x + 2)(x - 1)

Now for x² + 6x + 8. Two numbers that multiply to 8 and add to 6. 4 and 2. (x + 4)(x + 2)

And finally x² + x - 12. Two numbers that multiply to -12 and add to 1 would be 4 and -3. (x + 4)(x - 3).

Put it all together...

[(x + 3)(x - 3)]/[(x + 2)(x - 1)] × [(x + 4)(x - 2)]/[(x + 4)(x - 3)]

When multiplying two fractions, you can multiply straight across, numerator times numerator and denominator times denominator.

[(x + 3)(x - 3)(x + 4)(x - 2)]/[(x + 2)(x - 1)(x + 4)(x - 3)]

Cancel out anything that is both in the numerator and denominator. (x - 3) cancels, and (x + 4) and (x + 2).

What's left is:

(x + 3)/(x - 1)

Answer 2

1)

Looks like x² was cancelled out of the numerator and denominator. Then 2 was divided out of the x-term of the numerator and denominator and then an 8 was divided out of the constant term of the numerator and denominator.

Instead, the fraction should have been factored first:

(x+4)(x-2)/((x+8)(x-2))

Then (x-2) can be cancelled for the numerator and denominator giving:

(x+4)/(x+8)

2)

First, flip the divisor fraction and then factor the numerators and denominators of both fractions.

(x+3)(x-3)/((x+2)(x-1)) • (x+4)(x+2)/((x+4)(x-3))

Now, the (x-3)'s, (x+4)'s and (x+2)'s can all be cancelled, leaving

(x+3)/(x-1)