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Common Core: High School - Algebra : Identify Zeros, Factor and Graph Polynomials: CCSS.Math.Content.HSA-APR.B.3

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

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #151 : Arithmetic With Polynomials & Rational Expressions

What are the -intercept(s) of the function?

y=x^2+12x+36

Possible Answers:

$\text{x-intercept: }6$

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+12x+36

the coefficients are,

$\\b=12 \\c=36$

therefore the factors of that have a sum of are,

y=(x+6)(x+6)

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

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

To verify, graph the function.

Screen shot 2016 03 09 at 10.18.10 am

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

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Example Question #152 : Arithmetic With Polynomials & Rational Expressions

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

Possible Answers:

-35 , 35

$\sqrt{1493} , - \sqrt{1493}$

There are no real roots

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

Correct answer:

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

Explanation:

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

Recall the quadratic formula.

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

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

$a x^{2} + b x + c = 0$

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

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

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

Now we split this up into two equations.

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

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

So our zeros are at

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

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

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Example Question #153 : Arithmetic With Polynomials & Rational Expressions

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

Possible Answers:

$\sqrt{1493} , - \sqrt{1493}$

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

There are no real roots

Correct answer:

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

Explanation:

$a x^{2} + b x + c = 0$

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

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

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

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

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

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

Now we split this up into two equations.

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

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

So our zeros are at

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

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

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Example Question #154 : Arithmetic With Polynomials & Rational Expressions

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

Possible Answers:

-14 , 14

There are no real roots

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

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

Correct answer:

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

Explanation:

$a x^{2} + b x + c = 0$

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

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

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

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

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

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

Now we split this up into two equations.

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

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

So our zeros are at

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

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

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Example Question #231 : High School: Algebra

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

Possible Answers:

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

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

There are no real roots

-34 , 34

Correct answer:

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

Explanation:

$a x^{2} + b x + c = 0$

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

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

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

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

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

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

Now we split this up into two equations.

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

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

So our zeros are at

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

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

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

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

Possible Answers:

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

There are no real roots

-15 , 15

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

Correct answer:

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

Explanation:

$a x^{2} + b x + c = 0$

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

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

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

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

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

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

Now we split this up into two equations.

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

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

So our zeros are at

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

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

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Example Question #233 : High School: Algebra

Find the zeros of $- 5 x^{2} + 36 x + 76$

Possible Answers:

-36 , 36

There are no real roots

$\frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

$16 \sqrt{11} , - 16 \sqrt{11}$

Correct answer:

$\frac{18}{5} + \frac{8 \sqrt{11}}{5} , - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

Explanation:

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

Recall the quadratic formula.

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

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

$a x^{2} + b x + c = 0$

In this case a = -5 , b = 36 and c = 76

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

$x =\frac{ -36 \pm \sqrt{ 36 ^2 - 4 * -5 * 76 }}{ 2 * -5 }$

$x =\frac{ -36 \pm \sqrt{ 1296 - -1520 }}{ -10 }$

$x =\frac{ -36 \pm \sqrt{ 2816 }}{ -10 }$

$x =\frac{ -36 \pm 16 \sqrt{11} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -36 + 16 \sqrt{11} }{ -10 }$

$x =\frac{ -36 - 16 \sqrt{11} }{ -10 }$

So our zeros are at

$x = \frac{18}{5} + \frac{8 \sqrt{11}}{5}$

$x = - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

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Example Question #234 : High School: Algebra

Find the zeros of $4 x^{2} + 29 x + 19$

Possible Answers:

$\sqrt{537} , - \sqrt{537}$

There are no real roots

$- \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$

-29 , 29

Correct answer:

$- \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$

Explanation:

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

Recall the quadratic formula.

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

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

$a x^{2} + b x + c = 0$

In this case a = 4 , b = 29 , and c = 19

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

$x =\frac{ -29 \pm \sqrt{ 29 ^2 - 4 * 4 * 19 }}{ 2 * 4 }$

$x =\frac{ -29 \pm \sqrt{ 841 - 304 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{ 537 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{537} }{ 8 }$

Now we split this up into two equations.

$x =\frac{ -29 + \sqrt{537} }{ 8 }$

$x =\frac{ -29 - \sqrt{537} }{ 8 }$

So our zeros are at

$x = - \frac{29}{8} - \frac{\sqrt{537}}{8}$

$x = - \frac{29}{8} + \frac{\sqrt{537}}{8}$

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Example Question #235 : High School: Algebra

Find the zeros of $x^{2} + 34 x + 24$

Possible Answers:

$-17 - \sqrt{265} , -17 + \sqrt{265}$

-34 , 34

$2 \sqrt{265} , - 2 \sqrt{265}$

There are no real roots

Correct answer:

$-17 - \sqrt{265} , -17 + \sqrt{265}$

Explanation:

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 34 and c = 24

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

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 1 * 24 }}{ 2 * 1 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - 96 }}{ 2 }$

$x =\frac{ -34 \pm \sqrt{ 1060 }}{ 2 }$

$x =\frac{ -34 \pm 2 \sqrt{265} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -34 + 2 \sqrt{265} }{ 2 }$

$x =\frac{ -34 - 2 \sqrt{265} }{ 2 }$

So our zeros are at

$\\x = -17 - \sqrt{265} \\x = -17 + \sqrt{265}$

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Example Question #236 : High School: Algebra

Find the zeros of $- 5 x^{2} + 30 x - 16$

Possible Answers:

$- \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$

-30 , 30

There are no real roots

$2 \sqrt{145} , - 2 \sqrt{145}$

Correct answer:

$- \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$

Explanation:

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

Recall the quadratic formula.

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

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

$a x^{2} + b x + c = 0$

In this case a = -5 , b = 30 and c = -16

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

$x =\frac{ -30 \pm \sqrt{ 30 ^2 - 4 * -5 * -16 }}{ 2 * -5 }$

$x =\frac{ -30 \pm \sqrt{ 900 - 320 }}{ -10 }$

$x =\frac{ -30 \pm \sqrt{ 580 }}{ -10 }$

$x =\frac{ -30 \pm 2 \sqrt{145} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -30 + 2 \sqrt{145} }{ -10 }$

$x =\frac{ -30 - 2 \sqrt{145} }{ -10 }$

So our zeros are at

$x = - \frac{\sqrt{145}}{5} + 3$

$x = \frac{\sqrt{145}}{5} + 3$

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Common Core: High School - Algebra : Identify Zeros, Factor and Graph Polynomials: CCSS.Math.Content.HSA-APR.B.3

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #151 : Arithmetic With Polynomials & Rational Expressions

Example Question #152 : Arithmetic With Polynomials & Rational Expressions

Example Question #153 : Arithmetic With Polynomials & Rational Expressions

Example Question #154 : Arithmetic With Polynomials & Rational Expressions

Example Question #231 : High School: Algebra

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

Example Question #233 : High School: Algebra

Example Question #234 : High School: Algebra

Example Question #235 : High School: Algebra

Example Question #236 : High School: Algebra

All Common Core: High School - Algebra Resources

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