Intermediate Single-Variable Algebra - High School Math

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Question

$f(x)=x^{2}-4x+7$

Find the vertex of the parabola by completing the square.

Answer

To find the vertex of a parabola, we must put the equation into the vertex form:

$\small f(x)=a[b(x-h)]^{2}+k$

The vertex can then be found with the coordinates (h, k).

To put the parabola's equation into vertex form, you have to complete the square. Completing the square just means adding the same number to both sides of the equation -- which, remember, doesn't change the value of the equation -- in order to create a perfect square.

Start with the original equation:

$f(x)=x^{2}-4x+7$

Put all of the terms on one side:

$f(x)-7=x^{2}-4x$

Now we know that we have to add something to both sides in order to create a perfect square:

$f(x)+7+\square =x^{2}-4x+\square$

In this case, we need to add 4 on both sides so that the right-hand side of the equation factors neatly.

$f(x)-7+4=x^{2}-4x+4$

Now we factor:

$f(x)-3=(x-2)^{2}$

Once we isolate , we have the equation in vertex form:

$f(x)=(x-2)^{2}+3$

Thus, the parabola's vertex can be found at $\small (2,3)$ .

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Question

Complete the square:

Answer

Begin by dividing the equation by and subtracting from each side:

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

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

$x = \frac{-3 \pm i\sqrt{7}}{4}$

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Question

Use factoring to solve the quadratic equation:

Answer

Factor and solve:

Factor like terms:

Combine like terms:

$x=-\frac{3}{2},\frac{1}{3}$

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Question

Complete the square:

Answer

Begin by dividing the equation by and adding to each side:

$n^2+\frac{1}{2}n = \frac{21}{2}$

Square the value in front of the and add to each side:

$n^2+\frac{1}{2}n + \frac{1}{16} = \frac{21}{2} + \frac{1}{16}$

Factor the left side of the equation:

$(n+\frac{1}{4})^2 = \frac{169}{16}$

Take the square root of both sides and simplify:

$n+\frac{1}{4} = \frac{\pm 13}{4}$

$n = \frac{-1 \pm 13}{4}$

$n=-\frac{7}{2},3$

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Question

Complete the square:

Answer

Begin by dividing the equation by and subtracting $\frac{2}{3}$ from each side:

$x^2+\frac{4}{3}x = -\frac{2}{3}$

Square the value in front of the and add to each side:

$x^2+\frac{4}{3}x +\frac{4}{9} = -\frac{2}{3}+\frac{4}{9}$

Factor the left side of the equation:

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

Take the square root of both sides and simplify:

$x+\frac{2}{3} = \frac{\pm \sqrt{-2}}{3}$

$x = \frac{-2 \pm i\sqrt{2}}{3}$

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Question

Find the zeros.

$f(x)=x^{3}-1$

Answer

This is a difference of perfect cubes so it factors to . Only the first expression will yield an answer when set equal to 0, which is 1. The second expression will never cross the -axis. Therefore, your answer is only 1.

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Question

Find the zeros.

Answer

Factor the equation to . Set and get one of your 's to be . Then factor the second expression to . Set them equal to zero and you get $\pm \sqrt{2}$ .

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Question

Factor $x^{3}-64$

Answer

Use the difference of perfect cubes equation:

In $x^{3}-64$ ,

and $b=\sqrt[3]{64}=4$

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Question

Factor the polynomial completely and solve for .

Answer

To factor and solve for in the equation

Factor out of the equation

$\rightarrow 8x(4x^2-1)$

Use the "difference of squares" technique to factor the parenthetical term, which provides the completely factored equation:

$\rightarrow 8x(2x+1)(2x-1)$

Any value that causes any one of the three terms , , and to be will be a solution to the equation, therefore

$x = \left {-\frac{1}{2}, 0, \frac{1}{2} \right }$

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Question

Factor the following expression:

Answer

You can see that each term in the equation has an "x", therefore by factoring "x" from each term you can get that the equation equals .

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Question

Factor this expression:

Answer

First consider all the factors of 12:

1 and 12

2 and 6

3 and 4

Then consider which of these pairs adds up to 7. This pair is 3 and 4.

Therefore the answer is .

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Question

Factor the following polynomial:

Answer

Begin by extracting from the polynomial:

Now, factor the remainder of the polynomial as a difference of cubes:

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Question

Factor the following polynomial:

Answer

Begin by rearranging like terms:

Now, factor out like terms:

Rearrange the polynomial:

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Question

Factor the following polynomial:

Answer

Begin by rearranging like terms:

Now, factor out like terms:

Rearrange the polynomial:

Factor:

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Question

Factor the following polynomial:

Answer

Begin by separating into like terms. You do this by multiplying and , then finding factors which sum to

Now, extract like terms:

Simplify:

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Question

Factor the following polynomial:

Answer

To begin, distribute the squares:

Now, combine like terms:

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Question

Factor the following polynomial:

Answer

Begin by extracting from the polynomial:

Now, distribute the cubic polynomial:

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Question

Factor the following polynomial:

Answer

Begin by extracting like terms:

Now, rearrange the right side of the polynomial by reversing the signs:

Combine like terms:

Factor the square and cubic polynomial:

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Question

Factor the following polynomial:

Answer

Begin by rearranging the terms to group together the quadratic:

Now, convert the quadratic into a square:

Finally, distribute the :

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Question

Factor the following polynomial:

Answer

Begin by extracting from the polynomial:

Now, rearrange to combine like terms:

Extract the like terms and factor the cubic:

Simplify by combining like terms:

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