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College Algebra : Review and Other Topics

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Example Questions

Example Question #15 : Rational Expressions

Simplify: $\frac{3x^5}{2x^5+x}$

Possible Answers:

$\textup{The rational expression cannot be factored.}$

$x+\frac{3}{2}$

$\frac{3}{2}$

$\frac{3}{x+2}$

$\frac{3x^4}{2x^4+1}$

Correct answer:

$\frac{3x^4}{2x^4+1}$

Explanation:

Factor an in the denominator.

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

Cancel the common in the numerator and denominator.

The answer is: $\frac{3x^4}{2x^4+1}$

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Example Question #391 : College Algebra

Add: $\frac{1}{x}+\frac{2}{x-1}$

Possible Answers:

$\frac{3x-1}{x^2-1}$

$-\frac{x-1}{x^2-x}$

$\frac{3x-1}{x^2-x}$

$\frac{3}{x^2-x}$

$-\frac{x-3}{x^2-x}$

Correct answer:

$\frac{3x-1}{x^2-x}$

Explanation:

To add the terms of the numerator, we will need to have the same denominator.

Multiply both denominators together.

x(x-1) = x^2-x

Multiply the numerators with what was multiplied on the denominator.

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

Combine like terms.

The answer is: $\frac{3x-1}{x^2-x}$

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Example Question #391 : College Algebra

Add/Subtract: $\frac {4}{5}-\frac {1}{4}$ $\frac {4}{5}-\frac {1}{4}+1$

Possible Answers:

None of the Above

$1\frac {11}{20}$

Correct answer:

$1\frac {11}{20}$

Explanation:

Step 1: Find the Least Common denominator of all the fractions..

LCD= 4*5*1=20

Step 2: Convert all fractions to denominator .

$\frac {4}{5}\equiv \frac {16}{20}$

$\frac {1}{4}\equiv \frac {5}{20}$

$1=\frac {20}{20}$

Step 3: Re-write all fractions in terms of the original equation...

$\frac {16}{20}-\frac {5}{20}+\frac {20}{20}$

Answer: $\frac {31}{20}$ or $1\frac {11}{20}$

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Example Question #1 : Basic Operations With Complex Numbers

Consider the following definitions of imaginary numbers:

x = 4 - 2i

y = 6 +7i

z = 5i

Then, x+y-2z = ?

Possible Answers:

-5i

10-i

2(5+i)

5(2-i)

Correct answer:

5(2-i)

Explanation:

x + y-2z = (4-2i) + (6+7i) -2(5i) = 10 - 5i = 5(2-i)

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Example Question #1 : Basic Operations With Complex Numbers

What is the value of (3 + i)(4 + i) ?

Possible Answers:

11 + 7i

12 + 8i

12 + 6i

Correct answer:

11 + 7i

Explanation:

When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below:

(3+i)(4+i)= 12 + 3i + 4i + i^2

Since we know that i^2 = -1 we get 12 + 7i - 1 which gives us 11 + 7i .

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Example Question #1 : Basic Operations With Complex Numbers

What is the value of i^4 ?

Possible Answers:

i + 1

Correct answer:

Explanation:

Recall that the definition of imaginary numbers gives that $i = \sqrt{-1}$ and thus that i^2 = -1 . Therefore, we can use Exponent Rules to write $i^4 = i^2 \cdot i^2 = -1\cdot-1 = 1$

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Example Question #6 : How To Add Integers

Add:

(2-3i)+(-1-2i)

Possible Answers:

1-5i

1+5i

5-3i

5+3i

-1+5i

Correct answer:

1-5i

Explanation:

When adding complex numbers, add the real parts and the imaginary parts separately to get another complex number in standard form.

Adding the real parts gives 2-1=1 , and adding the imaginary parts gives -5i .

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Example Question #2 : Complex Numbers

Divide: $\frac{2-i}{3+2i}$

The answer must be in standard form.

Possible Answers:

$\frac{2}{3}i$

$\frac{2}{3}-\frac{1}{2}i$

$\frac{4}{13}-\frac{7i}{13}$

$\frac{1}{5i}$

Correct answer:

$\frac{4}{13}-\frac{7i}{13}$

Explanation:

Multiply both the numerator and the denominator by the conjugate of the denominator which is 3-2i which results in

$\frac{\left ( 2-i \right )\left ( 3-2i \right )}{\left ( 3+2i \right )\left ( 3-2i \right )}$

The numerator after simplification give us $6-4i-3i+2i^{2}=6-7i+2i^{2}=4-7i$

The denominator is equal to $3^{2}-4i^{2}=9+4=13$

Hence, the final answer in standard form =

$\frac{4}{13}-\frac{7i}{13}$

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Example Question #2 : Complex Numbers

Divide: $\frac{2+3i}{i}$

Answer must be in standard form.

Possible Answers:

3+2i

3-2i

2i-3

Correct answer:

3-2i

Explanation:

Multiply both the numerator and the denominator by the conjugate of the denominator which is resulting in

$\frac{\left ( 2+3i \right )\left ( -i \right )}{\left ( i \right )\left ( -i \right )}$

This is equal to $\frac{-2i-{3i}^2}{-i^2}$

Since i^2 = -1 you can make that substitution of in place of i^2 in both numerator and denominator, leaving:

$\frac{-2i-3(-1)}{-(-1)}$

When you then cancel the negatives in both numerator and denominator (remember that -(-1)=1 , simplifying each term), you're left with a denominator of and a numerator of -2i + 3 , which equals 3-2i .

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Example Question #21 : Sat Subject Test In Math I

Evaluate: i(1+i)(2-i)

Possible Answers:

-1-3i

-1+3i

1+3 i

-3+i

1-3i

Correct answer:

-1+3i

Explanation:

Use the FOIL method to simplify. FOIL means to mulitply the first terms together, then multiply the outer terms together, then multiply the inner terms togethers, and lastly, mulitply the last terms together.

i(1+i)(2-i)

=i[(1)(2)+(1)(-i)+(i)(2)+(i)(-i)]

=i[2-i+2i-i^2]

=i[2+i-i^2]

=2i+i^2-i^3

The imaginary is equal to:

$\sqrt{-1}$

Write the terms for $i, i^2, \textup{and } i^3$ .

$i=\sqrt{-1}$

i^2=-1

$i^3=i \times i^2=-\sqrt{-1}=-i$

Replace $i^2 \textup{ and } i^3$ with the appropiate values and simplify.

2i+i^2-i^3= 2i-1-(-i) = -1+3i

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College Algebra : Review and Other Topics

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #15 : Rational Expressions

Example Question #391 : College Algebra

Example Question #391 : College Algebra

Example Question #1 : Basic Operations With Complex Numbers

Example Question #1 : Basic Operations With Complex Numbers

Example Question #1 : Basic Operations With Complex Numbers

Example Question #6 : How To Add Integers

Example Question #2 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #21 : Sat Subject Test In Math I

All College Algebra Resources

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