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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #2 : Factoring Rational Expressions

Simplify.

$\frac{2x^2-4x-6}{4x-12}$

Possible Answers:

$\frac{x+1}{2}$

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

The expression cannot be simplified.

x+1

2x+2

Correct answer:

$\frac{x+1}{2}$

Explanation:

a. Simplify the numerator and denominator separately by pulling out common factors.

$\frac{2(x^2-2x-3)}{4(x-3)}$

b. Reduce if possible.

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

c. Factor the trinomial in the numerator.

$\frac{(x-3)(x+1)}{2(x-3)}$

d. Cancel common factors between the numerator and the denominator.

$\frac{{\color{Red} (x-3)}(x+1)}{2{\color{Red} (x-3)}}=\frac{x+1}{2}$

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Example Question #1791 : Algebra Ii

Transform the following equation from standard into vertex form:

y=3x^2+18x+15

Possible Answers:

y=(9x+2)^2-12

y=(3x+8)^2

y=3(x+3)^2+15

y=3(x+3)^2-12

y=(3x+3)^2-18

Correct answer:

y=3(x+3)^2-12

Explanation:

To take this standard form equation and transform it into vertex form, we need to complete the square. That can be done as follows:

$\small y=3x^2+18x+15$

y-15=3(x^2+6x)

We will complete the square on 3(x^2+6x) . In this case, our in our soon-to-be ax^2+bx+c is . We therefore want our $c=(\frac{1}{2}b)^2$ , so $\small c=(\frac{1}{2}(6))^2=9$ .

Since we are adding $\small 3\times9$ on the right side (as we are completing the square inside the parentheses), we need to add $\small 27$ on the left side as well. Our equation therefore becomes:

$\small y-15+27=3(x^2+6x+9)$

$\small y+12=3(x+3)^2$

Our final answer is therefore $\small y=3(x+3)^2-12$

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Example Question #4 : Factoring Rational Expressions

Evaluate the following expression: $\small \small \frac{2x^2y^8}{5z^4}*\frac{10x^5z^5}{4y^6}$

Possible Answers:

$\small \frac{2y^1^4z^9}{5x^3}$

$\small x^7y^2z$

$\small \frac{1}{\small x^7y^2z}$

$\small \frac{y^1^4z^9}{x^3}$

$\small \small \frac{5x^3}{2y^1^4z^9}$

Correct answer:

$\small x^7y^2z$

Explanation:

When we multiply expressions with exponents, we need to keep in mind some rules:

Multiplied variables add exponents.

Divided variables subtract exponents.

Variables raised to a power multiply exponents.

Therefore, when we mulitiply the two fractions, we obtain:

$\small \frac{2x^2y^8}{5z^4}\ast\frac{10x^5z^5}{4y^6}=\frac{20x^7y^8z^5}{20y^6z^4}=x^7y^2z$

Our final answer is therefore $\small x^7y^2z$

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Example Question #1 : Factoring Rational Expressions

Simplify:

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

Possible Answers:

x+1

x-1

x+2

x-3

x+3

Correct answer:

x+1

Explanation:

First factor the numerator. We need two numbers with a sum of 3 and a product of 2. The numbers 1 and 2 satisfy these conditions:

(x+2)(x+1)

Now, look to see if there are any common factors that will cancel:

$\frac{(x+2)(x+1)}{x+2}$

The x+2 in the numerator and denominator cancel, leaving x+1 .

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Example Question #1 : Factoring Rational Expressions

Simplify this rational expression: $\frac{x^2-14x+49}{x^2-2x-35}\cdot \frac{1}{x+5}$

Possible Answers:

$\frac{(x-7)^2}{x+5}$

$\frac{(x-7)^2}{(x+5)^2}$

$\frac{x-7}{(x+5)^2}$

None of the other answers.

$\frac{x-7}{x+5}$

Correct answer:

$\frac{x-7}{(x+5)^2}$

Explanation:

To see what can be simplified, factor the quadratic equations.

$\frac{(x-7)(x-7)}{(x-7)(x+5)}\cdot \frac{1}{x+5}$

Cancel out like terms:

$\frac{{\color{Red} (x-7)}(x-7)}{{\color{Red} (x-7)}(x+5)}\cdot \frac{1}{x+5}\rightarrow \frac{x-7}{x+5}\cdot \frac{1}{x+5}$

Combine terms:

$\frac{x-7}{(x+5)^2}$

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Example Question #7 : Factoring Rational Expressions

Factor and simplify this rational expression: $\frac{(x+2)^2(x-3)(x+3)}{(x^2+2x+4)(x^2-6x+9)(x^2+6x+9)}$

Possible Answers:

None of these.

$\frac{1}{(x+3)(x+3)}$

$\frac{1}{(x-3)(x+3)}$

$\frac{1}{(x+3)}$

$\frac{1}{x^2+6x+9}$

Correct answer:

$\frac{1}{(x-3)(x+3)}$

Explanation:

$\frac{(x+2)^2(x-3)(x+3)}{(x^2+2x+4)(x^2-6x+9)(x^2+6x+9)}$

Completely factor all polynomials:

$\frac{(x+2)^2(x-3)(x+3)}{(x+2)(x+2)(x-3)(x-3)(x+3)(x+3)}$

Cancel like terms:

$\mathbf{\frac{1}{(x-3)(x+3)}}$

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Example Question #8 : Factoring Rational Expressions

Factor $f(x)=\frac{x^2 + x - 12}{x^2 - x - 6}$ .

Possible Answers:

(x+4)(x+2)

x(x+4)

$\frac{(x+3)(x+4)}{(x-3)(x+2)}$

$\frac{(x+4)}{(x+2)}$

$\frac{(x-3)}{(x+3)(x+2)}$

Correct answer:

$\frac{(x+4)}{(x+2)}$

Explanation:

In the beginning, we can treat this as two separate problems, and factor the numerator and the denominator independently:

x^2 + x - 12 = (x+4)(x-3)

x^2 - x - 6 = (x+2)(x-3)

After we've factored them, we can put the factored equations back into the original problem:

$f(x)= \frac{(x+4)(x-3)}{(x+2)(x-3)}$

From here, we can cancel the (x-3) from the top and the bottom, leaving:

$f(x)=\frac{(x+4)}{(x+2)}$

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Example Question #9 : Factoring Rational Expressions

Factor: $\frac{2x^2+8x-10}{16x-16}$

Possible Answers:

$\frac{x+5}{8}$

$\frac{x+3}{16}$

$\frac{x+5}{16}$

$\frac{x^2-4x-5}{8x}$

$\frac{x^2-4x-5}{8}$

Correct answer:

$\frac{x+5}{8}$

Explanation:

Factor a two out in the numerator.

2x^2+8x-10 = 2(x^2+4x-5)

Factor the trinomial.

2(x^2+4x-5) = 2(x-1)(x+5)

Factor the denominator.

16x-16 = 16(x-1)

Divide the terms.

$\frac{2(x-1)(x+5)}{16(x-1)}$

The answer is: $\frac{x+5}{8}$

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Example Question #2 : Factoring Rational Expressions

Simplify to simplest terms.

$\frac{x^2-x-56}{x^2-15x+56}$

Possible Answers:

$\frac{x+7}{x-7}$

Correct answer:

Explanation:

The correct answer is . The numerator and denominator can both be factored to simpler terms:

$\frac{(x-8)(x+7)}{(x-8)(x-7)}$

The (x-8 ) terms will cancel out. Leaving $\frac{x+7}{x-7}$ . While this is an answer choice, it can be simplified further. Factoring out a from the denominator will allow the x-7 terms to cancel out leaving .

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Example Question #1791 : Algebra Ii

Simplify the rational expression by factoring:

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

Possible Answers:

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

None of these.

$\frac{(x+2)^3}{(x+3)}$

$\frac{(x+2)^2}{(x+3)}$

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

Correct answer:

$\frac{(x+2)^3}{(x+3)}$

Explanation:

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

To simplify it is best to completely factor all polynomials:

$\frac{(x+2)(x+2)(x+2)^2}{(x+2)(x+3)}$

Now cancel like terms:

$\frac{(x+2)(x+2)^2}{(x+3)}$

Combine like terms:

$\mathbf{\frac{(x+2)^3}{(x+3)}}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2 : Factoring Rational Expressions

Example Question #1791 : Algebra Ii

Example Question #4 : Factoring Rational Expressions

Example Question #1 : Factoring Rational Expressions

Example Question #1 : Factoring Rational Expressions

Example Question #7 : Factoring Rational Expressions

Example Question #8 : Factoring Rational Expressions

Example Question #9 : Factoring Rational Expressions

Example Question #2 : Factoring Rational Expressions

Example Question #1791 : Algebra Ii

All Algebra II Resources

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