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College Algebra : Zeros/Roots of a Polynomial

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Example Question #51 : Polynomial Functions

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

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 #51 : Polynomial Functions

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

Possible Answers:

x=-160,624

x=160,-624

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

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

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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Example Question #231 : College Algebra

$\begin{align*}&\text{Find the root(s) the function: }f(x)=71x - 153x^{2} - 6\end{align*}$

Possible Answers:

$x=\frac{1}{9},\frac{6}{17}$

x=34,108

x=-34,-108

$x=-\frac{1}{9},-\frac{6}{17}$

Correct answer:

$x=\frac{1}{9},\frac{6}{17}$

Explanation:

$\begin{align*}&\text{The roots of the function }f(x)=71x - 153x^{2} - 6\text{ are the x values}\\&\text{that give it a value of zero. To find these x values}\\&\text{we can either factor the equation:}\\&-(9x - 1)\cdot (17x - 6)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(71)\pm\sqrt{(71)^2-4(-153)(-6)}}{2(-153)}=\frac{-71\pm37}{-306}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{1}{9}\text{ and }\frac{6}{17}.\end{align*}$

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Example Question #31 : Zeros/Roots Of A Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=224 - 580x^{2} - 328x\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{2}{5},\frac{28}{29}$

x=1120,-464

$x=\frac{2}{5},-\frac{28}{29}$

x=-1120,464

Correct answer:

$x=\frac{2}{5},-\frac{28}{29}$

Explanation:

$\begin{align*}&\text{Given }f(x)=224 - 580x^{2} - 328x\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&-4\cdot (29x + 28)\cdot (5x - 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-328)\pm\sqrt{(-328)^2-4(-580)(224)}}{2(-580)}=\frac{328\pm792}{-1160}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{2}{5}\text{ and }-\frac{28}{29}.\end{align*}$

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Example Question #233 : College Algebra

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 176x - 28x^{2} - 48\\&\text{crosses the x-axis.}\end{align*}$

Possible Answers:

$x=6,\frac{2}{7}$

$x=-6,-\frac{2}{7}$

x=-336,-16

x=336,16

Correct answer:

$x=-6,-\frac{2}{7}$

Explanation:

$\begin{align*}&\text{The function }f(x)=- 176x - 28x^{2} - 48\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-4\cdot (x + 6)\cdot (7x + 2)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-176)\pm\sqrt{(-176)^2-4(-28)(-48)}}{2(-28)}=\frac{176\pm160}{-56}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=-6\text{ and }-\frac{2}{7}.\end{align*}$

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Example Question #52 : Polynomial Functions

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=208x^{2} - 6x - 154\\&\text{has a zero value.}\end{align*}$

Possible Answers:

$x=-\frac{7}{8},\frac{11}{13}$

x=364,-352

$x=\frac{7}{8},-\frac{11}{13}$

x=-364,352

Correct answer:

$x=\frac{7}{8},-\frac{11}{13}$

Explanation:

$\begin{align*}&\text{Given }f(x)=208x^{2} - 6x - 154\text{, to find when it has a value of}\\&\text{zero, we need a method to determine the x values that}\\&\text{make this happen. We can either factor the equation:}\\&2\cdot (13x + 11)\cdot (8x - 7)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(-6)\pm\sqrt{(-6)^2-4(208)(-154)}}{2(208)}=\frac{6\pm358}{416}\\&\text{Either way, we find that the function crosses the x-axis at}\ x=\frac{7}{8}\text{ and }-\frac{11}{13}.\end{align*}$

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Example Question #231 : College Algebra

Consider the polynomial

$p(x) = x^{5}+ 5x^{3}- 7x +1$

Which of the following is true of the rational zeroes of ?

Hint: Think "Rational Zeroes Theorem".

Possible Answers:

There are no rational zeroes.

is the only rational zero.

1 is the only rational zero.

Neither nor 1 is a rational zero, but there is at least one rational zero.

and 1 are the only rational zeroes.

Correct answer:

1 is the only rational zero.

Explanation:

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set $\left \{ -1, 1\right \}$ .

1 is a zero of if and only if p(1) =0 . An easy test for this is to add the coefficients and determine whether their sum, which is p(1) , is 0:

$p(x) = x^{5}+ 5x^{3}- 7x +1$

p(1) = 1 + 5-7+1

p(1)= 0

1 is a zero.

is a zero of if and only if p(-1) =0 . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is p(1) , is 0:

$p(x) = x^{5}+ 5x^{3}- 7x +1$

p(-1) = -1-5+7+1

$p(-1) = 2 \ne 0$

is not a zero.

1 is the only rational zero.

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Example Question #236 : College Algebra

The polynomial

$P(x)= 2x^{5}+16x^{4}-40x^{3}-3x^{2}-24x+60$

has a zero on one of the following intervals. Which one?

Possible Answers:

(1.2, 1.3)

(1.3, 1.4)

( 1.1, 1.2 )

(1.4, 1.5)

(1.0, 1.1)

Correct answer:

( 1.1, 1.2 )

Explanation:

The Intermediate value Theorem states that if P(x) is a continuous function, a< b , and P(a) and P(b) are of unlike sign, it must hold that P(c)= 0 for some $c \in [a,b]$ . Equivalently, if P(x) has values on the endpoints of an interval that differ in sign, then P(x) has a zero on that interval. Since all polynomial functions are continuous, we may apply this theorem by evaluating $P(x)= 2x^{5}+16x^{4}-40x^{3}-3x^{2}-24x+60$

for x = 1.0, 1.1, 1.2, 1.3, 1.4, 1.5 , and finding two consecutive values such that P(x) differs in sign.

By substitution, we can find that:

P(1.0)= 11

P(1.1)= 3.37662

P(1.2) = -4.08576

P(1.3) = -11.02654

P(1.4)=-17.01792

P(1.5) = -21.5625

The change in sign occurs on the interval ( 1.1, 1.2 ) ; therefore, a zero of is located on this interval.

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Example Question #232 : College Algebra

Consider the polynomial

$x^{5}+ 8 x^{3}- 42x + 17$ .

The Rational Zeroes Theorem allows us to reduce the possible rational zeroes of this polynomial to a set comprising how many elements?

Possible Answers:

Correct answer:

Explanation:

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

$\left \{ \pm 1, \pm 17 \right \}$ ,

a set of four elements. This makes 4 the correct choice.

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Example Question #61 : Polynomial Functions

Consider the polynomial

$x^{5}+ 8 x^{3}- 42x + 17$ .

Going by Descartes' Rule of Signs alone, how many positive real zeroes could the polynomial have?

Possible Answers:

One or three

Exactly two

Exactly three

None or two

Exactly one

Correct answer:

None or two

Explanation:

By Descartes' Rule of Signs, the number of positive real zeroes of a polynomial p(x) with real coefficients is, at maximum, the number of its sign changes, but equal to some number of the same odd-even parity. This number can be seen to be two:

$x^{5}+ 8 x^{3}\textbf{{\color{Red} - }}42x \textbf{{\color{Red} +}} 17$

This means that this polynomial must have either zero or two such zeroes.

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College Algebra : Zeros/Roots of a Polynomial

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #51 : Polynomial Functions

Example Question #51 : Polynomial Functions

Example Question #231 : College Algebra

Example Question #31 : Zeros/Roots Of A Polynomial

Example Question #233 : College Algebra

Example Question #52 : Polynomial Functions

Example Question #231 : College Algebra

Example Question #236 : College Algebra

Example Question #232 : College Algebra

Example Question #61 : Polynomial Functions

All College Algebra Resources

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