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College Algebra : Polynomial Functions

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Example Question #11 : Finding Zeros Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=286x^{2} - 43x - 204\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{12}{13},\frac{17}{22}$

$x=\frac{48}{13},-\frac{34}{11}$

$x=\frac{25081}{572},\frac{24111}{572}$

$x=\frac{12}{13},-\frac{17}{22}$

Correct answer:

$x=\frac{12}{13},-\frac{17}{22}$

Explanation:

$\begin{align*}&\text{The function }f(x)=286x^{2} - 43x - 204\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&(22x + 17)\cdot (13x - 12)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-43)\pm\sqrt{(-43)^2-4(286)(-204)}}{2(286)}=\frac{43\pm485}{572}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{12}{13}\text{ and }-\frac{17}{22}.\end{align*}$

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Example Question #11 : Finding Zeros Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=288x - 175x^{2} + 256\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{16}{7},\frac{16}{25}$

$x=-\frac{50656}{175},-\frac{50144}{175}$

$x=\frac{64}{7},-\frac{64}{25}$

$x=\frac{16}{7},-\frac{16}{25}$

Correct answer:

$x=\frac{16}{7},-\frac{16}{25}$

Explanation:

$\begin{align*}&\text{The function }f(x)=288x - 175x^{2} + 256\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-(25x + 16)\cdot (7x - 16)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(288)\pm\sqrt{(288)^2-4(-175)(256)}}{2(-175)}=\frac{-288\pm512}{-350}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{16}{7}\text{ and }-\frac{16}{25}.\end{align*}$

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Example Question #81 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 154x - 35x^{2} - 147\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=\frac{766}{5},\frac{774}{5}$

$x=-12,-\frac{28}{5}$

$x=-3,-\frac{7}{5}$

$x=3,\frac{7}{5}$

Correct answer:

$x=-3,-\frac{7}{5}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 154x - 35x^{2} - 147\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-7\cdot (x + 3)\cdot (5x + 7)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-154)\pm\sqrt{(-154)^2-4(-35)(-147)}}{2(-35)}=\frac{154\pm56}{-70}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-3\text{ and }-\frac{7}{5}.\end{align*}$

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Example Question #11 : Finding Zeros Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=15 - 76x^{2} - 281x\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=\frac{1}{19},-\frac{15}{4}$

$x=\frac{4}{19},-15$

$x=-\frac{1}{19},\frac{15}{4}$

$x=\frac{42423}{152},\frac{43001}{152}$

Correct answer:

$x=\frac{1}{19},-\frac{15}{4}$

Explanation:

$\begin{align*}&\text{The function }f(x)=15 - 76x^{2} - 281x\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-(4x + 15)\cdot (19x - 1)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-281)\pm\sqrt{(-281)^2-4(-76)(15)}}{2(-76)}=\frac{281\pm289}{-152}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=\frac{1}{19}\text{ and }-\frac{15}{4}.\end{align*}$

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Example Question #82 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 435x - 84x^{2} - 450\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=\frac{15}{4},\frac{10}{7}$

$x=-15,-\frac{40}{7}$

$x=\frac{24295}{56},\frac{24425}{56}$

$x=-\frac{15}{4},-\frac{10}{7}$

Correct answer:

$x=-\frac{15}{4},-\frac{10}{7}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 435x - 84x^{2} - 450\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-3\cdot (4x + 15)\cdot (7x + 10)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-435)\pm\sqrt{(-435)^2-4(-84)(-450)}}{2(-84)}=\frac{435\pm195}{-168}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{15}{4}\text{ and }-\frac{10}{7}.\end{align*}$

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Example Question #83 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=199x + 200x^{2} - 28\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{1}{8},\frac{28}{25}$

$x=\frac{1}{8},-\frac{28}{25}$

$x=\frac{1}{2},-\frac{112}{25}$

$x=-\frac{79351}{400},-\frac{79849}{400}$

Correct answer:

$x=\frac{1}{8},-\frac{28}{25}$

Explanation:

$\begin{align*}&\text{The function }f(x)=199x + 200x^{2} - 28\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&(25x + 28)\cdot (8x - 1)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(199)\pm\sqrt{(199)^2-4(200)(-28)}}{2(200)}=\frac{-199\pm249}{400}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{1}{8}\text{ and }-\frac{28}{25}.\end{align*}$

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Example Question #12 : Finding Zeros Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=348x + 96x^{2} + 270\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-10,-\frac{9}{2}$

$x=\frac{5}{2},\frac{9}{8}$

$x=-\frac{5}{2},-\frac{9}{8}$

$x=-\frac{5557}{16},-\frac{5579}{16}$

Correct answer:

$x=-\frac{5}{2},-\frac{9}{8}$

Explanation:

$\begin{align*}&\text{The function }f(x)=348x + 96x^{2} + 270\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&6\cdot (2x + 5)\cdot (8x + 9)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(348)\pm\sqrt{(348)^2-4(96)(270)}}{2(96)}=\frac{-348\pm132}{192}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=-\frac{5}{2}\text{ and }-\frac{9}{8}.\end{align*}$

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Example Question #11 : Finding Zeros Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=251x - 33x^{2} - 400\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=-\frac{16}{3},-\frac{25}{11}$

$x=\frac{16}{3},\frac{25}{11}$

$x=-\frac{16667}{66},-\frac{16465}{66}$

$x=\frac{64}{3},\frac{100}{11}$

Correct answer:

$x=\frac{16}{3},\frac{25}{11}$

Explanation:

$\begin{align*}&\text{The function }f(x)=251x - 33x^{2} - 400\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-(11x - 25)\cdot (3x - 16)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(251)\pm\sqrt{(251)^2-4(-33)(-400)}}{2(-33)}=\frac{-251\pm101}{-66}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{16}{3}\text{ and }\frac{25}{11}.\end{align*}$

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Example Question #89 : Polynomial Functions

Find the zeros of the given polynomial:

2x^2-2x-15=0

Possible Answers:

x=-3,-5

x=-5,3

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

x=3,-5

$x=\frac{1\pm \sqrt{31}}{2}$

Correct answer:

$x=\frac{1\pm \sqrt{31}}{2}$

Explanation:

To find the values for in which the polynomial equals , we first want to factor the equation:

2x^2-2x-15=0

$x=\frac{-(-2)\pm \sqrt{(-2)^2-4(2)(-15)}}{2(2)}$

$\\x=\frac{2\pm \sqrt{4+120}}{4}$

$x=\frac{2\pm \sqrt{124}}{4}$

$x=\frac{2\pm 2\sqrt{31}}{4}$

$x=\frac{1\pm \sqrt{31}}{2}$

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Example Question #81 : Polynomial Functions

Consider the polynomial

$P(x) = 3x^{5}+ 6x^{3} + 17x^{2} + 7x + 10$

By Descartes' Rule of Signs alone, how many positive real zeroes does P(x) have?

(Note: you are not being asked for the actual number of positive real zeroes.)

Possible Answers:

Five

One or three

One

Two or Zero

Zero

Correct answer:

Zero

Explanation:

By Descartes' Rule of Signs, the number of sign changes - changes from positive to negative coefficient signs - in P(x) gives the maximum number of positive real zeroes; the actual number of positive real zeroes must be that many or differ by an even number.

If the polynomial

$P(x) = 3x^{5}+ 6x^{3} + 17x^{2} + 7x + 10$

is examined, it can be seen that there are no changes in sign from term to term; all coefficients are positive. Therefore, there can be no positive real zeroes.

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College Algebra : Polynomial Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #11 : Finding Zeros Of A Polynomial

Example Question #11 : Finding Zeros Of A Polynomial

Example Question #81 : Polynomial Functions

Example Question #11 : Finding Zeros Of A Polynomial

Example Question #82 : Polynomial Functions

Example Question #83 : Polynomial Functions

Example Question #12 : Finding Zeros Of A Polynomial

Example Question #11 : Finding Zeros Of A Polynomial

Example Question #89 : Polynomial Functions

Example Question #81 : Polynomial Functions

All College Algebra Resources

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