Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

Study concepts, example questions & explanations for Common Core: High School - Algebra

varsity tutors app store varsity tutors android store

All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #7 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the \displaystyle x-intercept(s) of the function?

\displaystyle y=x^2+4x+3

Possible Answers:

\displaystyle \text{x-intercepts:}-3,-1

\displaystyle \text{x-intercepts: }3,1

\displaystyle \text{x-intercepts:}-3,1

\displaystyle \text{x-intercepts:}-3,-1,1

\displaystyle \text{x-intercepts: }3,-1

Correct answer:

\displaystyle \text{x-intercepts:}-3,-1

Explanation:

To find the \displaystyle x-intercept of a function, first recall that the \displaystyle x-intercept represents the points where the graph of the function crosses the \displaystyle x-axis. In other words where the function has a \displaystyle y value equal to zero.

One technique that can be used is factorization. In general form,

\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)

where,

\displaystyle c_1 and \displaystyle c_2 are factors of \displaystyle c and when added together results in \displaystyle b.

For the given function,

 \displaystyle y=x^2+4x+3

the coefficients are,

\displaystyle \\b=4 \\c=3

therefore the factors of \displaystyle c that have a sum of \displaystyle b are,

\displaystyle y=(x+1)(x+3)

Now find the \displaystyle x-intercepts of the function by setting each binomial equal to zero and solving for \displaystyle x.

\displaystyle \\x+1=0\rightarrow x=-1 \\x+3=0\rightarrow x=-3

To verify, graph the function.

Screen shot 2016 03 08 at 1.52.17 pm

The graph crosses the \displaystyle x-axis at -1 and -3, thus verifying the results found by factorization. 

Example Question #8 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the \displaystyle x-intercept(s) of the function?

\displaystyle y=x^2-2x+1

Possible Answers:

\displaystyle \text{x-intercepts: }1,-1

\displaystyle \text{x-intercept: }1

\displaystyle \text{x-intercepts: }0,1

\displaystyle \text{x-intercept: }-1

\displaystyle \text{x-intercept: }0

Correct answer:

\displaystyle \text{x-intercept: }1

Explanation:

To find the \displaystyle x-intercept of a function, first recall that the \displaystyle x-intercept represents the points where the graph of the function crosses the \displaystyle x-axis. In other words where the function has a \displaystyle y value equal to zero.

One technique that can be used is factorization. In general form,

\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)

where,

\displaystyle c_1 and \displaystyle c_2 are factors of \displaystyle c and when added together results in \displaystyle b.

For the given function,

 \displaystyle y=x^2-2x+1

the coefficients are,

\displaystyle \\b=-2 \\c=1

therefore the factors of \displaystyle c that have a sum of \displaystyle b are,

\displaystyle y=(x-1)(x-1)

Now find the \displaystyle x-intercepts of the function by setting each binomial equal to zero and solving for \displaystyle x.

\displaystyle \\x-1=0\rightarrow x=1

To verify, graph the function.

Screen shot 2016 03 09 at 9.54.14 am

The graph crosses the \displaystyle x-axis at 1, thus verifying the result found by factorization. 

Example Question #9 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the \displaystyle x-intercept(s) of the function?

\displaystyle y=x^2-8x+7

Possible Answers:

\displaystyle \text{x-intercepts: }-1,7

\displaystyle \text{x-intercepts: }1,8

\displaystyle \text{x-intercepts: }1,-7

\displaystyle \text{x-intercepts: }-1,-7

\displaystyle \text{x-intercepts: }1,7

Correct answer:

\displaystyle \text{x-intercepts: }1,7

Explanation:

To find the \displaystyle x-intercept of a function, first recall that the \displaystyle x-intercept represents the points where the graph of the function crosses the \displaystyle x-axis. In other words where the function has a \displaystyle y value equal to zero.

One technique that can be used is factorization. In general form,

\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)

where,

\displaystyle c_1 and \displaystyle c_2 are factors of \displaystyle c and when added together results in \displaystyle b.

For the given function,

 \displaystyle y=x^2-8x+7

the coefficients are,

\displaystyle \\b=-8 \\c=7

therefore the factors of \displaystyle c that have a sum of \displaystyle b are,

\displaystyle y=(x-1)(x-7)

Now find the \displaystyle x-intercepts of the function by setting each binomial equal to zero and solving for \displaystyle x.

\displaystyle \\x-1=0\rightarrow x=1 \\x-7=0\rightarrow x=7

To verify, graph the function.

Screen shot 2016 03 09 at 10.02.27 am

The graph crosses the \displaystyle x-axis at 1 and 7, thus verifying the results found by factorization. 

Example Question #10 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the \displaystyle x-intercept(s) of the function?

\displaystyle y=x^2-5x+6

Possible Answers:

\displaystyle \text{x-intercepts: }2,3

\displaystyle \text{x-intercepts: }2,-3

\displaystyle \text{x-intercepts: }-2,-3

\displaystyle \text{x-intercepts: }2,1

\displaystyle \text{x-intercepts: }-2,3

Correct answer:

\displaystyle \text{x-intercepts: }2,3

Explanation:

To find the \displaystyle x-intercept of a function, first recall that the \displaystyle x-intercept represents the points where the graph of the function crosses the \displaystyle x-axis. In other words where the function has a \displaystyle y value equal to zero.

One technique that can be used is factorization. In general form,

\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)

where,

\displaystyle c_1 and \displaystyle c_2 are factors of \displaystyle c and when added together results in \displaystyle b.

For the given function,

 \displaystyle y=x^2-5x+6

the coefficients are,

\displaystyle \\b=-5 \\c=6

therefore the factors of \displaystyle c that have a sum of \displaystyle b are,

\displaystyle y=(x-2)(x-3)

Now find the \displaystyle x-intercepts of the function by setting each binomial equal to zero and solving for \displaystyle x.

\displaystyle \\x-2=0\rightarrow x=2 \\x-3=0\rightarrow x=3

To verify, graph the function.

Screen shot 2016 03 09 at 10.10.58 am

The graph crosses the \displaystyle x-axis at 2 and 3, thus verifying the results found by factorization. 

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the \displaystyle x-intercept(s) of the function?

\displaystyle y=x^2+12x+36

Possible Answers:

\displaystyle \text{x-intercepts: }-6,-1

\displaystyle \text{x-intercepts: }-6,1

\displaystyle \text{x-intercepts: }-6,6

\displaystyle \text{x-intercept: }6

\displaystyle \text{x-intercept: }-6

Correct answer:

\displaystyle \text{x-intercept: }-6

Explanation:

To find the \displaystyle x-intercept of a function, first recall that the \displaystyle x-intercept represents the points where the graph of the function crosses the \displaystyle x-axis. In other words where the function has a \displaystyle y value equal to zero.

One technique that can be used is factorization. In general form,

\displaystyle y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)

where,

\displaystyle c_1 and \displaystyle c_2 are factors of \displaystyle c and when added together results in \displaystyle b.

For the given function,

 \displaystyle y=x^2+12x+36

the coefficients are,

\displaystyle \\b=12 \\c=36

therefore the factors of \displaystyle c that have a sum of \displaystyle b are,

\displaystyle y=(x+6)(x+6)

Now find the \displaystyle x-intercepts of the function by setting each binomial equal to zero and solving for \displaystyle x.

\displaystyle \\x+6=0\rightarrow x=-6

To verify, graph the function.

Screen shot 2016 03 09 at 10.18.10 am

The graph crosses the \displaystyle x-axis at -6, thus verifying the result found by factorization. 

Example Question #222 : High School: Algebra

Find the zeros of \displaystyle x^{2} + 35 x - 67

Possible Answers:

\displaystyle \sqrt{1493} , - \sqrt{1493}

\displaystyle \frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}

There are no real roots


\displaystyle -35 , 35

\displaystyle 0


Correct answer:

\displaystyle \frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }

Where \displaystyle \uptext{a}, \displaystyle \uptext{b}, and \displaystyle \uptext{c}, correspond to the coefficients in the equation.


\displaystyle a x^{2} + b x + c = 0

In this case \displaystyle a = 1, \displaystyle b = 35 and \displaystyle c = -67

We plug in these values into the quadratic formula, and evaluate them.

\displaystyle x =\frac{ -35 \pm \sqrt{ 35 ^2 - 4 * 1 * -67 }}{ 2 * 1 }
\displaystyle x =\frac{ -35 \pm \sqrt{ 1225 - -268 }}{ 2 }
\displaystyle x =\frac{ -35 \pm \sqrt{ 1493 }}{ 2 }
\displaystyle x =\frac{ -35 \pm \sqrt{1493} }{ 2 }

Now we split this up into two equations.

\displaystyle x =\frac{ -35 + \sqrt{1493} }{ 2 }

\displaystyle x =\frac{ -35 - \sqrt{1493} }{ 2 }

So our zeros are at

\displaystyle x = - \frac{35}{2} + \frac{\sqrt{1493}}{2}

\displaystyle x = - \frac{\sqrt{1493}}{2} - \frac{35}{2}


Example Question #223 : High School: Algebra

Find the zeros of \displaystyle x^{2} + 35 x - 67

Possible Answers:

There are no real roots


\displaystyle \frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}

\displaystyle \sqrt{1493} , - \sqrt{1493}

Correct answer:

\displaystyle \frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }

Where \displaystyle \uptext{a}, \displaystyle \uptext{b}, and \displaystyle \uptext{c}, correspond to the coefficients in the equation.


\displaystyle a x^{2} + b x + c = 0

In this case \displaystyle a = 1, \displaystyle b = 35 and \displaystyle c = -67

We plug in these values into the quadratic formula, and evaluate them.

\displaystyle x =\frac{ -35 \pm \sqrt{ 35 ^2 - 4 * 1 * -67 }}{ 2 * 1 }


\displaystyle x =\frac{ -35 \pm \sqrt{ 1225 - -268 }}{ 2 }


\displaystyle x =\frac{ -35 \pm \sqrt{ 1493 }}{ 2 }


\displaystyle x =\frac{ -35 \pm \sqrt{1493} }{ 2 }

Now we split this up into two equations.

\displaystyle x =\frac{ -35 + \sqrt{1493} }{ 2 }

\displaystyle x =\frac{ -35 - \sqrt{1493} }{ 2 }

So our zeros are at

\displaystyle x = - \frac{35}{2} + \frac{\sqrt{1493}}{2}

\displaystyle x = - \frac{\sqrt{1493}}{2} - \frac{35}{2}


Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Find the zeros of \displaystyle - 7 x^{2} + 14 x + 33

Possible Answers:

\displaystyle 1 + \frac{2 \sqrt{70}}{7} , - \frac{2 \sqrt{70}}{7} + 1

\displaystyle -14 , 14

\displaystyle 0

\displaystyle 4 \sqrt{70} , - 4 \sqrt{70}

There are no real roots

Correct answer:

\displaystyle 1 + \frac{2 \sqrt{70}}{7} , - \frac{2 \sqrt{70}}{7} + 1

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }

Where \displaystyle \uptext{a}, \displaystyle \uptext{b}, and \displaystyle \uptext{c} correspond to the coefficients in the equation

\displaystyle a x^{2} + b x + c = 0

In this case \displaystyle a = -7,  \displaystyle b = 14 and \displaystyle c = 33

We plug in these values into the quadratic formula, and evaluate them.

\displaystyle x =\frac{ -14 \pm \sqrt{ 14 ^2 - 4 * -7 * 33 }}{ 2 * -7 }

\displaystyle x =\frac{ -14 \pm \sqrt{ 196 - -924 }}{ -14 }

\displaystyle x =\frac{ -14 \pm \sqrt{ 1120 }}{ -14 }

\displaystyle x =\frac{ -14 \pm 4 \sqrt{70} }{ -14 }

Now we split this up into two equations.

\displaystyle x =\frac{ -14 + 4 \sqrt{70} }{ -14 }

\displaystyle x =\frac{ -14 - 4 \sqrt{70} }{ -14 }

So our zeros are at

\displaystyle x = 1 + \frac{2 \sqrt{70}}{7}

\displaystyle x = - \frac{2 \sqrt{70}}{7} + 1

Example Question #12 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Find the zeros of \displaystyle 9 x^{2} + 34 x - 99

Possible Answers:

\displaystyle 4 \sqrt{295} , - 4 \sqrt{295}

There are no real roots

\displaystyle 0

\displaystyle -34 , 34

\displaystyle - \frac{17}{9} + \frac{2 \sqrt{295}}{9} , - \frac{2 \sqrt{295}}{9} - \frac{17}{9}

Correct answer:

\displaystyle - \frac{17}{9} + \frac{2 \sqrt{295}}{9} , - \frac{2 \sqrt{295}}{9} - \frac{17}{9}

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }

Where \displaystyle \uptext{a}, \displaystyle \uptext{b}, and \displaystyle \uptext{c}, correspond to the coefficients in the equation


\displaystyle a x^{2} + b x + c = 0

In this case \displaystyle a = 9\displaystyle b = 34 and \displaystyle c = -99

We plug in these values into the quadratic formula, and evaluate them.

\displaystyle x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 9 * -99 }}{ 2 * 9 }


\displaystyle x =\frac{ -34 \pm \sqrt{ 1156 - -3564 }}{ 18 }

\displaystyle x =\frac{ -34 \pm \sqrt{ 4720 }}{ 18 }

\displaystyle x =\frac{ -34 \pm 4 \sqrt{295} }{ 18 }

Now we split this up into two equations.

\displaystyle x =\frac{ -34 + 4 \sqrt{295} }{ 18 }


\displaystyle x =\frac{ -34 - 4 \sqrt{295} }{ 18 }

So our zeros are at

\displaystyle x = - \frac{17}{9} + \frac{2 \sqrt{295}}{9}

\displaystyle x = - \frac{2 \sqrt{295}}{9} - \frac{17}{9}

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Find the zeros of \displaystyle 9 x^{2} + 15 x - 52

Possible Answers:

\displaystyle 3 \sqrt{233} , - 3 \sqrt{233}

There are no real roots

\displaystyle -15 , 15

\displaystyle - \frac{5}{6} + \frac{\sqrt{233}}{6} , - \frac{\sqrt{233}}{6} - \frac{5}{6}

\displaystyle 0

Correct answer:

\displaystyle - \frac{5}{6} + \frac{\sqrt{233}}{6} , - \frac{\sqrt{233}}{6} - \frac{5}{6}

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

\displaystyle x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }

Where \displaystyle \uptext{a}, \displaystyle \uptext{b}, and \displaystyle \uptext{c}, correspond to the coefficients in the equation

\displaystyle a x^{2} + b x + c = 0

In this case \displaystyle a = 9\displaystyle b = 15 and \displaystyle c = -52

We plug in these values into the quadratic formula, and evaluate them.

\displaystyle x =\frac{ -15 \pm \sqrt{ 15 ^2 - 4 * 9 * -52 }}{ 2 * 9 }

\displaystyle x =\frac{ -15 \pm \sqrt{ 225 - -1872 }}{ 18 }


\displaystyle x =\frac{ -15 \pm \sqrt{ 2097 }}{ 18 }

\displaystyle x =\frac{ -15 \pm 3 \sqrt{233} }{ 18 }

Now we split this up into two equations.

\displaystyle x =\frac{ -15 + 3 \sqrt{233} }{ 18 }

\displaystyle x =\frac{ -15 - 3 \sqrt{233} }{ 18 }

So our zeros are at

\displaystyle x = - \frac{5}{6} + \frac{\sqrt{233}}{6}

\displaystyle x = - \frac{\sqrt{233}}{6} - \frac{5}{6}

All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept
Learning Tools by Varsity Tutors