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High School Math : Intermediate Single-Variable Algebra

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Example Questions

Example Question #1 : Using Foil

FOIL $\displaystyle (x+2)(x-2)$ .

Possible Answers:

x^2-4 $\displaystyle x^2-4$

$\displaystyle x^2+4x+4$

$\displaystyle x^2-2x+2$

x^2+x-2 $\displaystyle x^2+x-2$

x^2+4 $\displaystyle x^2+4$

Correct answer:

x^2-4 $\displaystyle x^2-4$

Explanation:

Remember FOIL stands for First Outer Inner Last. That means we can take $\displaystyle (x+2)(x-2)$ and turn it into $\displaystyle x^2+2x-2x-4$ .

Simplify to get x^2-4 $\displaystyle x^2-4$ .

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Example Question #42 : Intermediate Single Variable Algebra

FOIL (x+5)^2 $\displaystyle (x+5)^2$ .

Possible Answers:

$\displaystyle x^2-10x+25$

$\displaystyle x^2+10x+25$

$\displaystyle x^2+25x+25$

5x^2 $\displaystyle 5x^2$

x^2+25 $\displaystyle x^2+25$

Correct answer:

$\displaystyle x^2+10x+25$

Explanation:

(x+5)^2 $\displaystyle (x+5)^2$ is the same thing as $\displaystyle (x+5)(x+5)$ .

Remember that FOIL stands for First Outer Inner Last.

For this problem, that would be $\displaystyle x^2+5x+5x+25$ .

Simplify that to $\displaystyle x^2+10x+25$ .

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Example Question #3 : Understanding Quadratic Equations

FOIL $\displaystyle (x+5)(x-1)$ .

Possible Answers:

$\displaystyle x^2+4x-5$

x^2+24 $\displaystyle x^2+24$

$\displaystyle x^2-4x-5$

x^2-5 $\displaystyle x^2-5$

$\displaystyle x^2+5x+5$

Correct answer:

$\displaystyle x^2+4x-5$

Explanation:

Remember that FOIL stands for First Outer Inner Last.

For this problem that would give us:

$\displaystyle (x+5)(x-1)$

$\displaystyle x^2-x+5x-5$

Simplify.

$\displaystyle x^2+4x-5$

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Example Question #44 : Intermediate Single Variable Algebra

FOIL $\displaystyle (x+2)(x-2)$ .

Possible Answers:

x^2-4 $\displaystyle x^2-4$

x^2+4 $\displaystyle x^2+4$

$\displaystyle x^2-4x-4$

x^2-4x $\displaystyle x^2-4x$

-x^2+4 $\displaystyle -x^2+4$

Correct answer:

x^2-4 $\displaystyle x^2-4$

Explanation:

Remember that FOIL stands for First Outer Inner Last.

For this problem that would give us:

$\displaystyle (x+2)(x-2)$

$\displaystyle =x^2-2x+2x-4$

Simplify:

=x^2-4 $\displaystyle =x^2-4$

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Example Question #1 : Using Foil

Solve the equation for $\displaystyle x$ .

$\small \frac{1}{x}=\frac{x+1}{6}$ $\displaystyle \small \frac{1}{x}=\frac{x+1}{6}$

Possible Answers:

$\small x=3\ or\ -2$ $\displaystyle \small x=3\ or\ -2$

$\small x=-3\ or\ 2$ $\displaystyle \small x=-3\ or\ 2$

$\small x=3\ or\ 2$ $\displaystyle \small x=3\ or\ 2$

$\small x=-3\ or\ -2$ $\displaystyle \small x=-3\ or\ -2$

Correct answer:

$\small x=-3\ or\ 2$ $\displaystyle \small x=-3\ or\ 2$

Explanation:

$\small \small \frac{1}{x}=\frac{x+1}{6}$ $\displaystyle \small \small \frac{1}{x}=\frac{x+1}{6}$

Cross multiply.

$\small 6=x(x+1)$ $\displaystyle \small 6=x(x+1)$

$\small 6=x^2+x$ $\displaystyle \small 6=x^2+x$

Set the equation equal to zero.

$\small 0=x^2+x-6$ $\displaystyle \small 0=x^2+x-6$

Factor to find the roots of the polynomial.

3*-2=-6 $\displaystyle 3*-2=-6$ and $\displaystyle 3+(-2)=1$

$\small 0=(x+3)(x-2)$ $\displaystyle \small 0=(x+3)(x-2)$

$\small 0=x+3; x=-3$ $\displaystyle \small 0=x+3; x=-3$

$\small 0=x-2; x=2$ $\displaystyle \small 0=x-2; x=2$

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Example Question #2 : Using Foil

Evaluate $(2x+3)^{2}$ $\displaystyle (2x+3)^{2}$

Possible Answers:

$\dpi{100} 4x+12$ $\displaystyle \dpi{100} 4x+12$

$\dpi{100} 4x^{2}+9x+9$ $\displaystyle \dpi{100} 4x^{2}+9x+9$

$\dpi{100} 4x^{3}+12x^{2}+9x$ $\displaystyle \dpi{100} 4x^{3}+12x^{2}+9x$

$\dpi{100} \dpi{100} 4x^{2}+9$ $\displaystyle \dpi{100} \dpi{100} 4x^{2}+9$

$4x^{2}+12x+9$ $\displaystyle 4x^{2}+12x+9$

Correct answer:

$4x^{2}+12x+9$ $\displaystyle 4x^{2}+12x+9$

Explanation:

In order to evaluate $(2x+3)^{2}$ $\displaystyle (2x+3)^{2}$ one needs to multiply the expression by itself using the laws of FOIL. In the foil method, one multiplies in the following order: first terms, outer terms, inner terms, and last terms.

$\dpi{100} (2x+3)^{2}=(2x+3)(2x+3)$ $\displaystyle \dpi{100} (2x+3)^{2}=(2x+3)(2x+3)$

Multiply terms by way of FOIL method.

$\displaystyle =(2x*2x)+(2x*3)+(3*2x)+(3*3)$

Now multiply and simplify.

$=4x^{2}+6x+6x+9$ $\displaystyle =4x^{2}+6x+6x+9$

$\rightarrow 4x^{2}+12x+9$ $\displaystyle \rightarrow 4x^{2}+12x+9$

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Example Question #41 : Intermediate Single Variable Algebra

Expand $\displaystyle (x+7)(x-2)$ .

Possible Answers:

x^2+9 $\displaystyle x^2+9$

$\displaystyle x^2+7x-2x-7$

$\displaystyle x^2-5x-14$

$\displaystyle x^2+5x-14$

$\displaystyle x^2-14x+5$

Correct answer:

$\displaystyle x^2+5x-14$

Explanation:

To solve our given equation, we need to use FOIL (First, Outer, Inner, Last).

$\displaystyle (x+7)(x-2)=x^2-2x+7x-14$

Combine like terms.

$\displaystyle (x+7)(x-2)=x^2+5x-14$

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Example Question #42 : Intermediate Single Variable Algebra

FOIL $\displaystyle (x+1)(x-3)$ .

Possible Answers:

$\displaystyle x^2-4x-3$

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

$\displaystyle x^2+3x+4$

$\displaystyle x^2-2x-3$

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

Correct answer:

$\displaystyle x^2-2x-3$

Explanation:

Remember FOIL stands for First Outer Inner Last.

$\displaystyle (x+1)(x-3)=x^2+1x-3x-3$

Combine like terms to get $\displaystyle x^2-2x-3$ .

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Example Question #293 : Algebra Ii

Use the discriminant to determine the nature of the roots:

$\displaystyle 4x^2-40x+25=0$

Possible Answers:

$\displaystyle 2$ irrational roots

$\displaystyle 2$ rational roots

$\displaystyle 2$ imaginary roots

$\displaystyle 1$ imaginary root

$\displaystyle 1$ rational root

Correct answer:

$\displaystyle 2$ irrational roots

Explanation:

The formula for the discriminant is:

$\displaystyle b^2 - 4ac$

$\displaystyle =(-40)^2 - 4(4)(25)$

$=1200^{}$ $\displaystyle =1200^{}$

Since the discriminant is positive and not a perfect square, there are $\displaystyle 2$ irrational roots.

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Example Question #12 : Quadratic Equations And Inequalities

Use the discriminant to determine the nature of the roots:

$\displaystyle 2y^2+6y+5=0$

Possible Answers:

$\displaystyle 1$ rational root

$\displaystyle 1$ imaginary root

$\displaystyle 2$ imaginary roots

$\displaystyle 2$ irrational roots

$\displaystyle 2$ rational roots

Correct answer:

$\displaystyle 2$ imaginary roots

Explanation:

The formula for the discriminant is:

$\displaystyle b^2 - 4ac$

$\displaystyle =(6)^2 - 4(2)(5)$

=-4 $\displaystyle =-4$

Since the discriminant is negative, there are $\displaystyle 2$ imaginary roots.

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High School Math : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Using Foil

Example Question #42 : Intermediate Single Variable Algebra

Example Question #3 : Understanding Quadratic Equations

Example Question #44 : Intermediate Single Variable Algebra

Example Question #1 : Using Foil

Example Question #2 : Using Foil

Example Question #41 : Intermediate Single Variable Algebra

Example Question #42 : Intermediate Single Variable Algebra

Example Question #293 : Algebra Ii

Example Question #12 : Quadratic Equations And Inequalities

All High School Math Resources

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