College Algebra - Zeros/Roots of a Polynomial

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=323x + 29x^{2} - 300\&\text{has a zero value.}\end{align*}$

$x=-12,\frac{25}{29}$

$x=12,-\frac{25}{29}$

Explanation

$\begin{align*}&\text{Given }f(x)=323x + 29x^{2} - 300\text{, to find when it has a value of}\&\text{zero, we need a method to determine the x values that}\&\text{make this happen. We can either factor the equation:}\&(x + 12)\cdot (29x - 25)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(323)\pm\sqrt{(323)^2-4(29)(-300)}}{2(29)}=\frac{-323\pm373}{58}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-12\text{ and }\frac{25}{29}.\end{align*}$

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=225x^{2} - 623x + 374\&\text{crosses the x-axis.}\end{align*}$

$x=\frac{22}{25},\frac{17}{9}$

$x=-\frac{22}{25},-\frac{17}{9}$

Explanation

$\begin{align*}&\text{The function }f(x)=225x^{2} - 623x + 374\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&(25x - 22)\cdot (9x - 17)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(-623)\pm\sqrt{(-623)^2-4(225)(374)}}{2(225)}=\frac{623\pm227}{450}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{22}{25}\text{ and }\frac{17}{9}.\end{align*}$

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

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

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

Explanation

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

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

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

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

Explanation

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

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".

1 is the only rational zero.

and 1 are the only rational zeroes.

is the only rational zero.

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

There are no rational zeroes.

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 . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

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

1 is a zero.

is a zero of if and only if . 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 , is 0:

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

is not a zero.

1 is the only rational zero.

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?

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.

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

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

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

Explanation

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

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

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

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

Explanation

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

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

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

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

Explanation

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

$\begin{align*}&\text{Find the value(s) of x where the function: }f(x)=- 544x - 152x^{2} - 480\&\text{crosses the x-axis.}\end{align*}$

$x=-2,-\frac{30}{19}$

$x=2,\frac{30}{19}$

Explanation

$\begin{align*}&\text{The function }f(x)=- 544x - 152x^{2} - 480\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&-8\cdot (x + 2)\cdot (19x + 30)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(-544)\pm\sqrt{(-544)^2-4(-152)(-480)}}{2(-152)}=\frac{544\pm64}{-304}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-2\text{ and }-\frac{30}{19}.\end{align*}$

Zeros/Roots of a Polynomial

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