Finding Zeros of a Polynomial - College Algebra

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Find the roots of the function:
$\small $f(x)=x^2$+4x-12$

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Factor:

$\small \small 0=(x-2)(x+6)$

Double check by factoring:

$\small F: (x)(x)$

$\small \small O:(x)(6)=6x$

$\small \small I: (-2)(x)=-2x$

$\small \small L: (-2)(6)=-12$

Add together: $\small $x^2$$+6x-2x-12=x^2$+4x-12$

Therefore:

$\small x-2=0\ or x+6=0$

$\small x=2\or-6$

Solve for x.

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Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

Now pull out the common factor, the "(x-2)," from both terms.

Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

Solve for :

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To solve for , you need to isolate it to one side of the equation. You can subtract the from the right to the left. Then you can add the 6 from the right to the left:

Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6:

$(x \ \ \ \ \ \)(x \ \ \ \ \ \) = 0$

Finally, you set each binomial equal to 0 and solve for :

$x=2 \ \ \ \ \ \ x=4$

Find the roots of $\small $9x^2$ -121$ .

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If we recognize this as an expression with form $\small $a^2$ - $b^2$$ , with $\small a=3x$ and $\small b=121$ , we can solve this equation by factoring:

$\small $9x^2$-121=0$

$\small (3x+11)(3x-11)=0$

$\small 3x+11=0$ and $\small 3x-11=0$

$\small x=-$\frac{11}{3}$$ and $\small x=\frac{11}{3}$$

Find the zeros of the given polynomial:

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

$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}$

If the following is a zero of a polynomial, find another zero.

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When finding zeros of a polynomial, you must remember your rules. Without a function this may seem tricky, but remember that non-real solutions come in conjugate pairs. Conjugate pairs differ in the middle sign. Thus, our answer is:

$\begin{align}&$\text{Find the value(s) of x where the function: }$f(x)=38x + $208x^{2}$ - 126\&$\text{crosses the x-axis.}$\end{align}$

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$\begin{align}&$\text{The function }$f(x)=38x + $208x^{2}$ - 126$\text{ crosses the x-axis}$\&$\text{when it has a value of zero. To find the x values that}$\&$\text{satisfy this, we can either factor the equation:}$\&2\cdot (8x + 7)\cdot (13x - 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$$+bx+c)\&$\frac{-(38)\pm\sqrt{(38)^2$-4(208)(-126)}$}{2(208)}=\frac{-38\pm326}{416}$\&$\text{Either way, we find thatthe function crosses the x-axis at}$ \ x=-$\frac{7}{8}$$\text{ and }$$\frac{9}{13}$.\end{align}$

$\begin{align}&$\text{Find the value(s) of x where the function: }$f(x)=200x - $320x^{2}$ + 250\&$\text{crosses the x-axis.}$\end{align}$

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$\begin{align}&$\text{The function }$f(x)=200x - $320x^{2}$ + 250$\text{ crosses the x-axis}$\&$\text{when it has a value of zero. To find the x values that}$\&$\text{satisfy this, we can either factor the equation:}$\&-10\cdot (8x + 5)\cdot (4x - 5)\&$\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$$+bx+c)\&$\frac{-(200)\pm\sqrt{(200)^2$-4(-320)(250)}$}{2(-320)}=\frac{-200\pm600}{-640}$\&$\text{Either way, we find thatthe function crosses the x-axis at}$ x=\frac{5}{4}$$\text{ and }$-$\frac{5}{8}$.\end{align}$

$\begin{align}&$\text{Find the value(s) of x where the function: }$f(x)=732x - $96x^{2}$ - 702\&$\text{crosses the x-axis.}$\end{align}$

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$\begin{align}&$\text{The function }$f(x)=732x - $96x^{2}$ - 702$\text{ crosses the x-axis}$\&$\text{when it has a value of zero. To find the x values that}$\&$\text{satisfy this, we can either factor the equation:}$\&-6\cdot (8x - 9)\cdot (2x - 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$$+bx+c)\&$\frac{-(732)\pm\sqrt{(732)^2$-4(-96)(-702)}$}{2(-96)}=\frac{-732\pm516}{-192}$\&$\text{Either way, we find thatthe function crosses the x-axis at}$\ x=\frac{13}{2}$$\text{ and }$$\frac{9}{8}$.\end{align}$

$\begin{align}&$\text{Find the value(s) of x where the function: }$f(x)=524x + $272x^{2}$ + 84\&$\text{crosses the x-axis.}$\end{align}$

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$\begin{align}&$\text{The function }$f(x)=524x + $272x^{2}$ + 84$\text{ crosses the x-axis}$\&$\text{when it has a value of zero. To find the x values that}$\&$\text{satisfy this, we can either factor the equation:}$\&4\cdot (4x + 7)\cdot (17x + 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$$+bx+c)\&$\frac{-(524)\pm\sqrt{(524)^2$-4(272)(84)}$}{2(272)}=\frac{-524\pm428}{544}$\&$\text{Either way, we find thatthe function crosses the x-axis at}$\ x=-$\frac{7}{4}$$\text{ and }$-$\frac{3}{17}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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

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$\begin{align}&$\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}$.\end{align}$

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 have?

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

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By Descartes' Rule of Signs, the number of sign changes - changes from positive to negative coefficient signs - in 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.

Determine the zeros of the following equation:

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To determine the zeros of this equation, we will need to factorize the polynomial.

The only common factors of that will give us a middle term of negative by addition or subtraction is:

Set each binomial equal to zero and solve.

The zeros are: