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College Algebra : College Algebra

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Example Questions

Example Question #23 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 295x - 250x^{2} - 66\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{3}{10},-\frac{22}{25}$

x=-440,-150

x=440,150

$x=\frac{3}{10},\frac{22}{25}$

Correct answer:

$x=-\frac{3}{10},-\frac{22}{25}$

Explanation:

$\begin{align*}&\text{Given }f(x)=- 295x - 250x^{2} - 66\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-(25x + 22)\cdot (10x + 3)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-295)\pm\sqrt{(-295)^2-4(-250)(-66)}}{2(-250)}=\frac{295\pm145}{-500}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{3}{10}\text{ and }-\frac{22}{25}.\end{align*}$

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Example Question #24 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=15x + 90x^{2} - 5\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{1}{6},\frac{1}{3}$

x=-30,60

$x=\frac{1}{6},-\frac{1}{3}$

x=30,-60

Correct answer:

$x=\frac{1}{6},-\frac{1}{3}$

Explanation:

$\begin{align*}&\text{Given }f(x)=15x + 90x^{2} - 5\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&5\cdot (3x + 1)\cdot (6x - 1)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(15)\pm\sqrt{(15)^2-4(90)(-5)}}{2(90)}=\frac{-15\pm45}{180}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{1}{6}\text{ and }-\frac{1}{3}.\end{align*}$

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Example Question #25 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=279x - 51x^{2} + 162\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=6,-\frac{9}{17}$

$x=-6,\frac{9}{17}$

x=54,-612

x=-54,612

Correct answer:

$x=6,-\frac{9}{17}$

Explanation:

$\begin{align*}&\text{Given }f(x)=279x - 51x^{2} + 162\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-3\cdot (17x + 9)\cdot (x - 6)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(279)\pm\sqrt{(279)^2-4(-51)(162)}}{2(-51)}=\frac{-279\pm333}{-102}\\&\text{Either way, we find that the function crosses the x-axis at} \ x=6\text{ and }-\frac{9}{17}.\end{align*}$

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Example Question #26 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=153x - 380x^{2} - 7\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=\frac{1}{19},\frac{7}{20}$

x=40,266

$x=-\frac{1}{19},-\frac{7}{20}$

x=-40,-266

Correct answer:

$x=\frac{1}{19},\frac{7}{20}$

Explanation:

$\begin{align*}&\text{Given }f(x)=153x - 380x^{2} - 7\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-(19x - 1)\cdot (20x - 7)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(153)\pm\sqrt{(153)^2-4(-380)(-7)}}{2(-380)}=\frac{-153\pm113}{-760}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{1}{19}\text{ and }\frac{7}{20}.\end{align*}$

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Example Question #27 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=154x + 14x^{2} + 252\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-2,-9

x=2,9

x=-56,-252

x=56,252

Correct answer:

x=-2,-9

Explanation:

$\begin{align*}&\text{Given }f(x)=154x + 14x^{2} + 252\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&14\cdot (x + 9)\cdot (x + 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(154)\pm\sqrt{(154)^2-4(14)(252)}}{2(14)}=\frac{-154\pm98}{28}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=-2\text{ and }-9.\end{align*}$

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Example Question #28 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=46 - 6x^{2} - 65x\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{2}{3},\frac{23}{2}$

$x=\frac{2}{3},-\frac{23}{2}$

x=138,-8

x=-138,8

Correct answer:

$x=\frac{2}{3},-\frac{23}{2}$

Explanation:

$\begin{align*}&\text{Given }f(x)=46 - 6x^{2} - 65x\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-(2x + 23)\cdot (3x - 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-65)\pm\sqrt{(-65)^2-4(-6)(46)}}{2(-6)}=\frac{65\pm73}{-12}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{2}{3}\text{ and }-\frac{23}{2}.\end{align*}$

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Example Question #29 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=221x - 184x^{2} + 44\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=-64,506

$x=\frac{4}{23},-\frac{11}{8}$

$x=-\frac{4}{23},\frac{11}{8}$

x=64,-506

Correct answer:

$x=-\frac{4}{23},\frac{11}{8}$

Explanation:

$\begin{align*}&\text{Given }f(x)=221x - 184x^{2} + 44\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-(23x + 4)\cdot (8x - 11)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(221)\pm\sqrt{(221)^2-4(-184)(44)}}{2(-184)}=\frac{-221\pm285}{-368}\\&\text{Either way, we find that the function crosses the x-axis at} \ x=-\frac{4}{23}\text{ and }\frac{11}{8}.\end{align*}$

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Example Question #51 : Polynomial Functions

$\begin{align*}&\text{Find the root(s) the function: }f(x)=66x^{2} - 248x - 280\end{align*}$

Possible Answers:

$x=-\frac{14}{3},\frac{10}{11}$

x=-616,120

$x=\frac{14}{3},-\frac{10}{11}$

x=616,-120

Correct answer:

$x=\frac{14}{3},-\frac{10}{11}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=66x^{2} - 248x - 280\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&2\cdot (11x + 10)\cdot (3x - 14)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-248)\pm\sqrt{(-248)^2-4(66)(-280)}}{2(66)}=\frac{248\pm368}{132}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{14}{3}\text{ and }-\frac{10}{11}.\end{align*}$

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Example Question #31 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=306x - 270x^{2} - 72\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{1}{3},-\frac{4}{5}$

$x=\frac{1}{3},\frac{4}{5}$

x=180,432

x=-180,-432

Correct answer:

$x=\frac{1}{3},\frac{4}{5}$

Explanation:

$\begin{align*}&\text{The function }f(x)=306x - 270x^{2} - 72\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-18\cdot (3x - 1)\cdot (5x - 4)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(306)\pm\sqrt{(306)^2-4(-270)(-72)}}{2(-270)}=\frac{-306\pm126}{-540}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{1}{3}\text{ and }\frac{4}{5}.\end{align*}$

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Example Question #32 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=232x + 120x^{2} - 208\\&\text{has a zero value.}\end{align*}$

Possible Answers:

x=160,-624

$x=\frac{2}{3},-\frac{13}{5}$

$x=-\frac{2}{3},\frac{13}{5}$

x=-160,624

Correct answer:

$x=\frac{2}{3},-\frac{13}{5}$

Explanation:

$\begin{align*}&\text{Given }f(x)=232x + 120x^{2} - 208\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&8\cdot (5x + 13)\cdot (3x - 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(232)\pm\sqrt{(232)^2-4(120)(-208)}}{2(120)}=\frac{-232\pm392}{240}\\&\text{Either way, we find that the function crosses the x-axis at} \ x=\frac{2}{3}\text{ and }-\frac{13}{5}.\end{align*}$

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College Algebra : College Algebra

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #23 : Zeros/Roots Of A Polynomial

Example Question #24 : Zeros/Roots Of A Polynomial

Example Question #25 : Zeros/Roots Of A Polynomial

Example Question #26 : Zeros/Roots Of A Polynomial

Example Question #27 : Zeros/Roots Of A Polynomial

Example Question #28 : Zeros/Roots Of A Polynomial

Example Question #29 : Zeros/Roots Of A Polynomial

Example Question #51 : Polynomial Functions

Example Question #31 : Zeros/Roots Of A Polynomial

Example Question #32 : Zeros/Roots Of A Polynomial

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