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Algebra 1 : Variables

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Example Questions

Example Question #3 : How To Factor A Polynomial

Solve by using the quadratic formula:

$6x^{2}+x +15$

Possible Answers:

$-1 + \sqrt{359}i$

$-1 - \sqrt{359}i$

$\frac{1}{12} - \frac{\sqrt{359}}{12}$

$\frac{-1}{12}+ \frac{\sqrt{359}i}{12}$

$\frac{-1}{12} - \frac{\sqrt{359 }i}{12}$

$\frac{1}{12} + \frac{\sqrt{359}}{12}$

$\frac{-1}{12} + \frac{\sqrt{359}}{12}$

Correct answer:

$\frac{-1}{12}+ \frac{\sqrt{359}i}{12}$

$\frac{-1}{12} - \frac{\sqrt{359 }i}{12}$

Explanation:

For the quadeatic equation $a=6,b=1,and\ c=15$ . Applying these to the quadratic formula

$\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

we get $\frac{-1 \pm \sqrt{1-4(6)15)}}{2\cdot 6}$

resulting in

$\frac{-1 \pm \sqrt{-359}}{12} = \frac{-1 +\sqrt{359}i}{12}$

and $\frac{-1 - \sqrt{359}i}{12}$

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Example Question #4551 : Algebra 1

Find the axis of symmetry and the minimum/maximum value of the following quadratic equation in standard form:

$- x^{2} + 4x - 12$

Possible Answers:

x=2 , Maximum value of the function =-8

x=-2 , Minimum value of the function =-8

x=2 , Minimum value of the function =-8

x=2 , Minimum value of the function

x=-2 , Maximum value of the function =-8

Correct answer:

x=2 , Maximum value of the function =-8

Explanation:

If we convert the given quadratic equation from standard form to vertex form, we get:

$-\left ( x-2 \right )^{2} -8$

Hence the axis of symmetry is

x=2 and the minimum value at x=2 is ( this function is concave down)

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Example Question #5 : Solving By Factoring

Write the equation of a circle having (3, 4) as center and a radius of $\sqrt{5}$ .

Possible Answers:

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = \sqrt{5}$

$\left ( x-3 \right )^{2} + \left ( y+4 \right )^{2} =\sqrt{5}$

$\left ( x-3 \right )^{3} + \left ( y +3 \right )^{2} = 5$

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

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

Correct answer:

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

Explanation:

The center is located at (3,4) which means the standard equation of a circle which is:

$x^{2} +y^{2} = r^{2}$

becomes

$\left ( x-3 \right )^{2} + \left ( y-4 \right )^{2} = \sqrt{5}^{2}$

which equals to

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

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Example Question #1 : Factoring Polynomials

Factor:

$8x^{3} - 1$

Possible Answers:

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

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

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

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

$\left ( 2x-1 \right )^{3}$

Correct answer:

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

Explanation:

This is a difference of cubes.

$8,x^{3},\ and\ -1$ are all cubes. So the formula for factoring this expression is:

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

$8x^{3}-1=(2x-1)(4x^{2}+2x+1)$

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Example Question #6 : Factoring Polynomials

Equation of a circle in standard form:

$\left ( x-h \right )^{2} + \left ( y - k \right )^{2} = r^{2}$

with center $\left ( h,k \right )$ and radius equal to .

Find out the radius and center of circle from the given equation in expanded form:

$x^{2} +2x + y^{2} -4x = 6$

Possible Answers:

The center is $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$

Center : $\left ( 1,-2 \right )$ and the radius = $\sqrt{11}$

Center: $\left ( 1,2 \right )$ and the radius = $\sqrt{11}$

Center: $\left ( 1,-2 \right )$ and the radius = $\sqrt{6}$

Center: $\left ( -1,-2 \right )$ and the radius = $\sqrt{11}$

Correct answer:

The center is $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$

Explanation:

The given equation in standard form :

$\left ( x+1 \right )^{2} + \left ( y-2 \right )^{2} = 11$

Hence the center: $\left ( -1,2 \right )$ and the radius = $\sqrt{11}$ .

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Example Question #2 : How To Factor A Polynomial

Factor:

$9x^{2} - 4y^{2}$

Possible Answers:

$\left ( 3x-2y \right )\left ( 3x-2y \right )$

$\left ( 3x+2y \right )\left ( 3x+2y \right )$

$\left ( 3x + 2y \right )^{2}$

$\left ( 3x + 2y \right )\left ( 3x - 2y \right )$

$\left ( 3x - 2y \right )^{2}$

Correct answer:

$\left ( 3x + 2y \right )\left ( 3x - 2y \right )$

Explanation:

Difference of 2 squares is equal to the product of the sum and difference of two terms.

The two terms here are and and the product therefore equals to

$\left ( 3x-2y \right )\left ( 3x + 2y \right )$

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Example Question #3 : Factoring Polynomials

Simplify:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

By factoring both the numerator and the denominator we get the following:

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

If we simplify we get:

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

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Example Question #2 : Solving By Factoring

Simplify:

$\frac{2x-5}{x-3}\div \frac{5-2x}{x-3}$

Possible Answers:

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

$\frac{5-2x}{3-x}$

$\frac{x-3}{2x-5}$

Correct answer:

Explanation:

Change division into multiplication by the reciprocal which gives us the following

$\frac{\left ( 2x-5 \right )\left ( x-3 \right )}{\left ( x-3 \right )\left ( 5-2x \right )}$

Now $\left ( 5-2x \right )=-(2x-5)$

this results in the following:

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

Simplification gives us

$\frac{1}{-1}$ which equals

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Example Question #321 : Variables

Simplify $\frac{3x+4}{6x^{2}+5x-4}$ .

Possible Answers:

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

2x-1

$\frac{1}{3x+7}$

$\frac{1}{2x+3}$

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

Correct answer:

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

Explanation:

Here, we simply need to identify that the numerator, 3x+4 , is a factor of the denominator. Let's start by factoring $6x^{2}+5x-4$ . The reverse FOIL method shows us that (3x+4)(2x-1) multiplies to give us $6x^{2}+5x-4$ , so we can rewrite the fraction as $\frac{(3x+4)}{(3x+4)(2x-1)}$ . Canceling the common term gives us our answer of $\frac{1}{2x-1}$ .

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Example Question #12 : How To Factor A Polynomial

Factor the polynomial completely: $16x^{2} + 64$

Possible Answers:

16 (x+2) (x-2)

The polynomial cannot be factored further.

$16 (x+2)^{2}$

$16 (x^{2}+4)$

$16(x-2)^{2}$

Correct answer:

$16 (x^{2}+4)$

Explanation:

The coefficients 16 and 64 have greatest common factor 16; there is no variable that is shared by both terms. Therefore, we can distribute out 16:

$16x^{2} + 64 = 16 \cdot x^{2 } + 16 \cdot 4 = 16 (x^{2}} + 4)$

$x^{2} + 4$ cannot be factored further, so $16 (x^{2}+4)$ is as far as we can go.

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Algebra 1 : Variables

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #3 : How To Factor A Polynomial

Example Question #4551 : Algebra 1

Example Question #5 : Solving By Factoring

Example Question #1 : Factoring Polynomials

Example Question #6 : Factoring Polynomials

Example Question #2 : How To Factor A Polynomial

Example Question #3 : Factoring Polynomials

Example Question #2 : Solving By Factoring

Example Question #321 : Variables

Example Question #12 : How To Factor A Polynomial

All Algebra 1 Resources

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