Zeros/Roots of a Polynomial - College Algebra

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$\begin{align}&\text{Find the value(s) of x where the function: }f(x)=350x + 112x^{2} + 42\&\text{crosses the x-axis.}\end{align}$

Answer

$\begin{align}&\text{The function }f(x)=350x + 112x^{2} + 42\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&14\cdot (x + 3)\cdot (8x + 1)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(350)\pm\sqrt{(350)^2-4(112)(42)}}{2(112)}=\frac{-350\pm322}{224}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-3\text{ and }-\frac{1}{8}.\end{align}$

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Question

$\begin{align}&\text{Find the value(s) of x where the function: }f(x)=- 197x - 532x^{2} - 15\&\text{has a zero value.}\end{align}$

Answer

$\begin{align}&\text{Given }f(x)=- 197x - 532x^{2} - 15\text{, to find when it has a value of}\&\text{zero, we need a method to determine the x values that}\&\text{make this happen. We can either factor the equation:}\&-(19x + 5)\cdot (28x + 3)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(-197)\pm\sqrt{(-197)^2-4(-532)(-15)}}{2(-532)}=\frac{197\pm83}{-1064}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{5}{19}\text{ and }-\frac{3}{28}.\end{align}$

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Question

$\begin{align}&\text{Find the value(s) of x where the function: }f(x)=24 - 900x^{2} - 300x\&\text{has a zero value.}\end{align}$

Answer

$\begin{align}&\text{Given }f(x)=24 - 900x^{2} - 300x\text{, to find when it has a value of}\&\text{zero, we need a method to determine the x values that}\&\text{make this happen. We can either factor the equation:}\&-12\cdot (5x + 2)\cdot (15x - 1)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(-300)\pm\sqrt{(-300)^2-4(-900)(24)}}{2(-900)}=\frac{300\pm420}{-1800}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{2}{5}\text{ and }\frac{1}{15}.\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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