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Algebra II : Algebra II

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Example Questions

Example Question #3 : How To Multiply Trinomials

Evaluate the following:

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

Possible Answers:

2x^3+9x^2+2x+20

2x^3+9x^2+18x-20

2x^3+9x^2+2x-20

4x^3+6x^2+2x-20

Correct answer:

2x^3+9x^2+2x-20

Explanation:

When multiplying this trinomial by this binomial, you'll need to use a modified form of FOIL, by which every term in the binomial gets multiplied by every term in the trinomial. One way to do this is to use the grid method.

You can also solve it piece-by-piece the way it is set up. First, multiply each of the three terms in the trinomail by . Then multiply each of those three terms again, this time by .

$(x^2+2x-4) \times 2x = 2x^3 + 4x^2 - 8x$

$(x^2+2x-4) \times 5 = 5x^2 + 10x - 20$

Finally, you can combine like terms after this multiplication to get your final simplified answer:

2x^3 + 9x^2 +2x-20

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Example Question #1 : Binomials

Expand:

(x-9)(3x+2)

Possible Answers:

$3x^{2}-18x-25$

None of the other answers

$3x^{2}-18x-29$

$3x^{2}-29x-18$

$3x^{2}-25x-18$

Correct answer:

$3x^{2}-25x-18$

Explanation:

To multiple these binomials, you can use the FOIL method to multiply each of the expressions individually.This will give you

(x)(3x)+(x)(2)+(-9)(3x)+(-9)(2)

or $3x^{2}-25x-18$ .

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Example Question #4271 : Algebra 1

Evaluate the following:

$\left(\frac{1}{2}x^2 +2x - 4\right) + \left(\frac{3}{2}x^2 -5x -4\right)$

Possible Answers:

2x^2-3x-8

2x^2+3x+8

x^2+3x-8

2x^2-3x

Correct answer:

2x^2-3x-8

Explanation:

To add these two trinomials, you will first begin by combining like terms. You have two terms with x^2 , two terms with , and two terms with no variable. For the two fractions with x^2 , you can immediately add because they have common denominators:

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

$\frac{4}{2}x^2 -3x - 8$

2x^2 -3x - 8

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Example Question #11 : Quadratic Equations And Inequalities

Simplify.

$\small \small \frac{x^2+5x+6}{x^3+2x^2-16x-32}$

Possible Answers:

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

$\small \small \frac{x+3}{(x+4)(x+4)}$

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

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

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

Correct answer:

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

Explanation:

Factoring the expression gives $\small \frac{}{}$ $\small \small \frac{(x+3)(x+2)}{(x-4)(x+4)(x+2)}$ . Values that are in both the numerator and denominator can be cancelled. By cancelling $\small x+2$ , the expression becomes $\small \small \frac{(x+3)}{(x-4)(x+4)}$ .

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Example Question #12 : Quadratic Equations And Inequalities

Simplfy.

$\small \frac{x^2-2x-15}{x^2-3x-10}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

By factoring the equation you get $\small \small \small \frac{(x+3)(x-5)}{(x+2)(x-5)}$ . Values that are in both the numerator and denominator can be cancelled. Cancelling the $\small x-5$ values gives $\small \small \small \frac{x+3}{x+2}$ .

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Example Question #13 : Quadratic Equations And Inequalities

Expand.

$\small 2(x+2)(x+4)$

Possible Answers:

$\small x^2+4x+8$

$\small \small 2x^2+12x-16$

$\small 2x+12x+16$

$\small 2x^2+12x+16$

$\small \small x^2+4x-8$

Correct answer:

$\small 2x^2+12x+16$

Explanation:

By foiling the binomials, multiplying the firsts, then the outers, followed by the inners and lastly the lasts, the expression you get is:

$\small 2(x^2+4x+2x+8)$ .

However, the expression can not be considered simplified in this state.

Distributing the two and adding like terms gives $\small 2x^2+12x+16$ .

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Example Question #12 : Understanding Quadratic Equations

If (ax+b)(2ax+3b)=8x^2+30x+27 , what is the value of a-b ?

Possible Answers:

Correct answer:

Explanation:

Use the FOIL method to simplify the binomial.

(a+b)(c+d) = ac+ad+bc+bd

(ax)(2ax)+(ax)(3b)+(b)(2ax)+(b)(3b)

Simplify the terms.

2a^2x^2+5abx+3b^2=8x^2+30x+27

Notice that the coefficients can be aligned to the unknown variables. Solve for and .

$2a^2 = 8 \rightarrow a^2 = 4 \rightarrow a=2$

$3b^2 = 27\rightarrow b^2=9 \rightarrow b=3$

a-b = 2-3 = -1

The answer is:

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Example Question #171 : Intermediate Single Variable Algebra

Multiply: (x^2-x-1)(3x^2-x+3)

Possible Answers:

$\textup{The answer is not given.}$

3x^4-3x^3-x^2-2x-3

3x^4-3x^3-x^2+2x-3

3x^4-4x^3+x^2-2x-3

3x^4-4x^3+4x^2-4x-3

Correct answer:

3x^4-4x^3+x^2-2x-3

Explanation:

Multiply each term of the first trinomial by second trinomial.

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

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

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

Add and combine like-terms.

The answer is: 3x^4-4x^3+x^2-2x-3

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Example Question #13 : Understanding Quadratic Equations

Simplify the function, if possible: $\frac{8x-10+2x^2}{-16+16x}$

Possible Answers:

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

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

$\frac{1}{16}x+\frac{5}{16}$

$\frac{1}{8}x+\frac{5}{8}$

$\frac{1}{16}x-\frac{5}{16}$

Correct answer:

$\frac{1}{8}x+\frac{5}{8}$

Explanation:

The expression will need to be rearranged from highest to lowest powers in order to be simplified.

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

Factor a 2 in the numerator.

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

Factor the term in parentheses.

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

Factor the denominator.

16x-16 = 16(x-1)

Divide the numerator with the denominator.

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

The expression becomes:

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

The answer is: $\frac{1}{8}x+\frac{5}{8}$

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Example Question #17 : Quadratic Equations And Inequalities

Solve for x:

$\frac{6}{x}=\frac{x+2}{4}$

Possible Answers:

-12,2

12,-2

-6,4

-4,6

None of the other answers

Correct answer:

-6,4

Explanation:

The correct answer is x=-6 or x=4 . The first step of the problem is to cross multiply. This will give the following equation:

24=x^2+2x

After subtracting from each side the equation looks like:

0=x^2+2x-24

The expression on the right hand side can be factored into:

0=(x-4)(x+6 )

Both and satisfy the above equation and are therefore the correct answers.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3 : How To Multiply Trinomials

Example Question #1 : Binomials

Example Question #4271 : Algebra 1

Example Question #11 : Quadratic Equations And Inequalities

Example Question #12 : Quadratic Equations And Inequalities

Example Question #13 : Quadratic Equations And Inequalities

Example Question #12 : Understanding Quadratic Equations

Example Question #171 : Intermediate Single Variable Algebra

Example Question #13 : Understanding Quadratic Equations

Example Question #17 : Quadratic Equations And Inequalities

All Algebra II Resources

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