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College Algebra : College Algebra

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Example Questions

Example Question #22 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 648x - 152x^{2} - 576\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

x=-912,-384

x=912,384

$x=-3,-\frac{24}{19}$

$x=3,\frac{24}{19}$

Correct answer:

$x=-3,-\frac{24}{19}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 648x - 152x^{2} - 576\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-8\cdot (x + 3)\cdot (19x + 24)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-648)\pm\sqrt{(-648)^2-4(-152)(-576)}}{2(-152)}=\frac{648\pm264}{-304}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=-3\text{ and }-\frac{24}{19}.\end{align*}$

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Example Question #201 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=712x - 110x^{2} - 728\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{26}{5},-\frac{14}{11}$

x=-280,-1144

$x=\frac{26}{5},\frac{14}{11}$

x=280,1144

Correct answer:

$x=\frac{26}{5},\frac{14}{11}$

Explanation:

$\begin{align*}&\text{The function }f(x)=712x - 110x^{2} - 728\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-2\cdot (11x - 14)\cdot (5x - 26)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(712)\pm\sqrt{(712)^2-4(-110)(-728)}}{2(-110)}=\frac{-712\pm432}{-220}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{26}{5}\text{ and }\frac{14}{11}.\end{align*}$

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Example Question #1 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 544x - 152x^{2} - 480\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-2,-\frac{30}{19}$

x=-608,-480

$x=2,\frac{30}{19}$

x=608,480

Correct answer:

$x=-2,-\frac{30}{19}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 544x - 152x^{2} - 480\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-8\cdot (x + 2)\cdot (19x + 30)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-544)\pm\sqrt{(-544)^2-4(-152)(-480)}}{2(-152)}=\frac{544\pm64}{-304}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-2\text{ and }-\frac{30}{19}.\end{align*}$

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Example Question #203 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=225x^{2} - 623x + 374\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

x=-850,-396

$x=\frac{22}{25},\frac{17}{9}$

x=850,396

$x=-\frac{22}{25},-\frac{17}{9}$

Correct answer:

$x=\frac{22}{25},\frac{17}{9}$

Explanation:

$\begin{align*}&\text{The function }f(x)=225x^{2} - 623x + 374\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&(25x - 22)\cdot (9x - 17)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-623)\pm\sqrt{(-623)^2-4(225)(374)}}{2(225)}=\frac{623\pm227}{450}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{22}{25}\text{ and }\frac{17}{9}.\end{align*}$

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Example Question #1 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=108x^{2} - 60x - 25\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=180,-60

$x=\frac{5}{6},-\frac{5}{18}$

$x=-\frac{5}{6},\frac{5}{18}$

$x=-\frac{5}{6},-\frac{5}{18}$

Correct answer:

$x=\frac{5}{6},-\frac{5}{18}$

Explanation:

$\begin{align*}&\text{Given }f(x)=108x^{2} - 60x - 25\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&(18x + 5)\cdot (6x - 5)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-60)\pm\sqrt{(-60)^2-4(108)(-25)}}{2(108)}=\frac{60\pm120}{216}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{5}{6}\text{ and }-\frac{5}{18}.\end{align*}$

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Example Question #1 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=421x + 306x^{2} - 210\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=238,-1080

x=-238,1080

$x=\frac{7}{18},-\frac{30}{17}$

$x=-\frac{7}{18},\frac{30}{17}$

Correct answer:

$x=\frac{7}{18},-\frac{30}{17}$

Explanation:

$\begin{align*}&\text{Given }f(x)=421x + 306x^{2} - 210\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&(17x + 30)\cdot (18x - 7)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(421)\pm\sqrt{(421)^2-4(306)(-210)}}{2(306)}=\frac{-421\pm659}{612}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{7}{18}\text{ and }-\frac{30}{17}.\end{align*}$

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Example Question #25 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=323x + 29x^{2} - 300\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-50,696

x=50,-696

$x=12,-\frac{25}{29}$

$x=-12,\frac{25}{29}$

Correct answer:

$x=-12,\frac{25}{29}$

Explanation:

$\begin{align*}&\text{Given }f(x)=323x + 29x^{2} - 300\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&(x + 12)\cdot (29x - 25)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(323)\pm\sqrt{(323)^2-4(29)(-300)}}{2(29)}=\frac{-323\pm373}{58}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-12\text{ and }\frac{25}{29}.\end{align*}$

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Example Question #2 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the root(s) the function: }f(x)=379x + 276x^{2} - 696\end{align*}$

Possible Answers:

$x=-\frac{24}{23},\frac{29}{12}$

x=576,-1334

$x=\frac{24}{23},-\frac{29}{12}$

x=-576,1334

Correct answer:

$x=\frac{24}{23},-\frac{29}{12}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=379x + 276x^{2} - 696\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&(12x + 29)\cdot (23x - 24)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(379)\pm\sqrt{(379)^2-4(276)(-696)}}{2(276)}=\frac{-379\pm955}{552}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{24}{23}\text{ and }-\frac{29}{12}.\end{align*}$

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Example Question #31 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=360x^{2} - 510x + 130\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{1}{3},-\frac{13}{12}$

$x=\frac{1}{3},\frac{13}{12}$

x=-780,-240

x=780,240

Correct answer:

$x=\frac{1}{3},\frac{13}{12}$

Explanation:

$\begin{align*}&\text{Given }f(x)=360x^{2} - 510x + 130\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&10\cdot (3x - 1)\cdot (12x - 13)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-510)\pm\sqrt{(-510)^2-4(360)(130)}}{2(360)}=\frac{510\pm270}{720}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{1}{3}\text{ and }\frac{13}{12}.\end{align*}$

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Example Question #32 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=360x - 250x^{2} + 64\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-80,800

$x=-\frac{8}{5},\frac{4}{25}$

$x=\frac{8}{5},-\frac{4}{25}$

x=80,-800

Correct answer:

$x=\frac{8}{5},-\frac{4}{25}$

Explanation:

$\begin{align*}&\text{Given }f(x)=360x - 250x^{2} + 64\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-2\cdot (25x + 4)\cdot (5x - 8)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(360)\pm\sqrt{(360)^2-4(-250)(64)}}{2(-250)}=\frac{-360\pm440}{-500}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{8}{5}\text{ and }-\frac{4}{25}.\end{align*}$

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College Algebra : College Algebra

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #22 : Polynomial Functions

Example Question #201 : College Algebra

Example Question #1 : Zeros/Roots Of A Polynomial

Example Question #203 : College Algebra

Example Question #1 : Zeros/Roots Of A Polynomial

Example Question #1 : Zeros/Roots Of A Polynomial

Example Question #25 : Polynomial Functions

Example Question #2 : Zeros/Roots Of A Polynomial

Example Question #31 : Polynomial Functions

Example Question #32 : Polynomial Functions

All College Algebra Resources

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