Algebra II - Simplifying and Expanding Quadratics

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

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

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

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

None of the other answers

Explanation

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

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

Subtract:

Explanation

When subtracting trinomials, first distribute the negative sign to the expression being subtracted, and then remove the parentheses:

Next, identify and group the like terms in order to combine them: .

Evaluate the following:

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:

Multiply:

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

Explanation

Multiply each term of the first trinomial by second trinomial.

Add and combine like-terms.

The answer is:

Simplify.

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

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

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

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

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

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

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

If , what is the value of ?

Explanation

Use the FOIL method to simplify the binomial.

Simplify the terms.

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

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

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

The answer is:

Simplfy.

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

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

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

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

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

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

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

Expand.

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

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

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

$\small 2x+12x+16$

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

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

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

If you were to solve $x^{2}+4x-5=0$ by completing the square, which of the following equations in the form do you get as a result?

Explanation

When given a quadratic in the form and told to solve by completing the square, we start by subtracting from both sides. In this problem is equal to , so we start by subtracting from both sides:

To complete the square we want to add a number to each side which yields a polynomial on the left side of the equation that can be simplified into a squared binomial . This number is equal to $(\frac{b}{2})^2$ . In this problem is equal to , so:

$(\frac{b}{2})^2=(\frac{4}{2})^2=2^2=4$

We add to both sides of the equation:

We then factor the left side of the equation into binomial squared form and combine like terms on the right:

If you were to solve $x^{2}+8x+9=0$ by completing the square, which of the following equations in the form do you get as a result?

Explanation

When given a quadratic in the form and told to solve by completing the square, we start by subtracting from both sides. In this problem is equal to , so we start by subtracting from both sides:

To complete the square we want to add a number to each side which yields a polynomial on the left side of the equals sign that can be simplified into a squared binomial . This number is equal to $(\frac{b}{2})^2$ . In this problem is equal to , so: