Question

Morgan made a mistake when subtracting the rational expressions below 3t^2-4t+1/t+3-t^2+2t+2/t+3=2t^2-2t+3/t+3. What was Morgan's error?
A. Morgan forgot to combine only like terms
B. Morgan forgot to subtract the denominators as well as numerators.
C. Morgan forgot to cancel out the +3 in the numerator and denominator as her final step.
D. Morgan forgot to distribute the negative sign to two of the terms in the second expression.

Answer 1

Answer: Choice D.

Morgan forgot to distribute the negative sign to two of the terms in the second expression.

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Explanation:

Focus on the numerators.

We have (3t^2-4t+1) as the first numerator and we subtract off (t^2+2t+2) as the second numerator.

Morgan needs to simplify (3t^2-4t+1)-(t^2+2t+2) for the numerator.

Mistakenly, she had these steps

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2+2t+2 .... her mistake made here

(3t^2-t^2)+(-4t+2t)+(1+2)

2t^2-2t+3

All of this applies to the numerator. The denominator stays at t+3 the entire time. So effectively we can ignore it on a temporary basis.

Here's what Morgan should have for her steps when simplifying the numerator.

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2-2t-2 ..... distribute the negative

(3t^2-t^2)+(-4t-2t)+(1-2)

2t^2-6t-1

Note in the second step, the negative outside flips the sign of each term in the second parenthesis.

Therefore,

$\frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}\\\\\frac{(3t^2-4t+1)-(t^2+2t+2)}{t+3}\\\\\frac{3t^2-4t+1-t^2-2t-2}{t+3}\\\\\frac{2t^2-6t-1}{t+3}$

which means $\frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}=\frac{2t^2-6t-1}{t+3}, \ \ \text{ where } t \ne -3$

Side notes:

• The fractions can only be subtracted since the denominators are the same.
• We have $t \ne -3$ to avoid a division by zero error.
• Rational expressions are a fraction, or ratio, of two polynomials.