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College Algebra : Polynomial Functions

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Example Questions

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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Example Question #32 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=322x^{2} - 325x + 75\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=\frac{5}{14},\frac{15}{23}$

$x=-\frac{5}{14},-\frac{15}{23}$

x=420,230

x=-420,-230

Correct answer:

$x=\frac{5}{14},\frac{15}{23}$

Explanation:

$\begin{align*}&\text{Given }f(x)=322x^{2} - 325x + 75\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&(14x - 5)\cdot (23x - 15)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-325)\pm\sqrt{(-325)^2-4(322)(75)}}{2(322)}=\frac{325\pm95}{644}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{5}{14}\text{ and }\frac{15}{23}.\end{align*}$

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Example Question #211 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=140x^{2} - 548x - 168\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

x=-1176,80

$x=-\frac{2}{7},\frac{21}{5}$

x=1176,-80

$x=\frac{2}{7},-\frac{21}{5}$

Correct answer:

$x=-\frac{2}{7},\frac{21}{5}$

Explanation:

$\begin{align*}&\text{The function }f(x)=140x^{2} - 548x - 168\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&4\cdot (7x + 2)\cdot (5x - 21)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-548)\pm\sqrt{(-548)^2-4(140)(-168)}}{2(140)}=\frac{548\pm628}{280}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{2}{7}\text{ and }\frac{21}{5}.\end{align*}$

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Example Question #212 : College Algebra

$\begin{align*}&\text{Find the root(s) the function: }f(x)=270x^{2} - 596x - 448\end{align*}$

Possible Answers:

x=-1512,320

x=1512,-320

$x=\frac{14}{5},-\frac{16}{27}$

$x=-\frac{14}{5},\frac{16}{27}$

Correct answer:

$x=\frac{14}{5},-\frac{16}{27}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=270x^{2} - 596x - 448\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&2\cdot (27x + 16)\cdot (5x - 14)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-596)\pm\sqrt{(-596)^2-4(270)(-448)}}{2(270)}=\frac{596\pm916}{540}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{14}{5}\text{ and }-\frac{16}{27}.\end{align*}$

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Example Question #213 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=123x - 460x^{2} + 532\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{28}{23},\frac{19}{20}$

x=874,-1120

x=-874,1120

$x=\frac{28}{23},-\frac{19}{20}$

Correct answer:

$x=\frac{28}{23},-\frac{19}{20}$

Explanation:

$\begin{align*}&\text{Given }f(x)=123x - 460x^{2} + 532\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-(20x + 19)\cdot (23x - 28)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(123)\pm\sqrt{(123)^2-4(-460)(532)}}{2(-460)}=\frac{-123\pm997}{-920}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{28}{23}\text{ and }-\frac{19}{20}.\end{align*}$

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Example Question #214 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=100x^{2} - 627x - 324\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=\frac{27}{4},-\frac{12}{25}$

x=1350,-96

$x=-\frac{27}{4},\frac{12}{25}$

x=-1350,96

Correct answer:

$x=\frac{27}{4},-\frac{12}{25}$

Explanation:

$\begin{align*}&\text{The function }f(x)=100x^{2} - 627x - 324\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&(25x + 12)\cdot (4x - 27)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-627)\pm\sqrt{(-627)^2-4(100)(-324)}}{2(100)}=\frac{627\pm723}{200}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{27}{4}\text{ and }-\frac{12}{25}.\end{align*}$

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Example Question #215 : College Algebra

$\begin{align*}&\text{Find the root(s) the function: }f(x)=293x - 75x^{2} - 66\end{align*}$

Possible Answers:

x=-36,-550

$x=\frac{6}{25},\frac{11}{3}$

x=36,550

$x=-\frac{6}{25},-\frac{11}{3}$

Correct answer:

$x=\frac{6}{25},\frac{11}{3}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=293x - 75x^{2} - 66\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&-(25x - 6)\cdot (3x - 11)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(293)\pm\sqrt{(293)^2-4(-75)(-66)}}{2(-75)}=\frac{-293\pm257}{-150}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{6}{25}\text{ and }\frac{11}{3}.\end{align*}$

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Example Question #216 : College Algebra

$\begin{align*}&\text{Find the root(s) the function: }f(x)=771x - 504x^{2} - 276\end{align*}$

Possible Answers:

$x=\frac{4}{7},\frac{23}{24}$

$x=-\frac{4}{7},-\frac{23}{24}$

x=576,966

x=-576,-966

Correct answer:

$x=\frac{4}{7},\frac{23}{24}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=771x - 504x^{2} - 276\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&-3\cdot (7x - 4)\cdot (24x - 23)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(771)\pm\sqrt{(771)^2-4(-504)(-276)}}{2(-504)}=\frac{-771\pm195}{-1008}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{4}{7}\text{ and }\frac{23}{24}.\end{align*}$

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College Algebra : Polynomial Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #2 : Zeros/Roots Of A Polynomial

Example Question #31 : Polynomial Functions

Example Question #32 : Polynomial Functions

Example Question #32 : Polynomial Functions

Example Question #211 : College Algebra

Example Question #212 : College Algebra

Example Question #213 : College Algebra

Example Question #214 : College Algebra

Example Question #215 : College Algebra

Example Question #216 : College Algebra

All College Algebra Resources

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