Finding Zeros of a Polynomial - College Algebra

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Question

Solve for x.

$x^{2}-7x+10=0$

Answer

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!

$x^{2}-2x-5x+10=0$

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

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Question

Solve for :

Answer

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 :

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

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Question

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

Answer

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:

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

$\small x=2\or-6$

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Question

Find the roots of the following quadratic expression:

Answer

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

$x = \frac{-4}{3}$

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

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Question

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

Answer

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

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Question

Find the zeros of the given polynomial:

Answer

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

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

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

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

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

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

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Question

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

Answer

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:

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Question

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

Answer

$\begin{align}&\text{The function }f(x)=38x + 208x^{2} - 126\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&2\cdot (8x + 7)\cdot (13x - 9)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(38)\pm\sqrt{(38)^2-4(208)(-126)}}{2(208)}=\frac{-38\pm326}{416}\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=-\frac{7}{8}\text{ and }\frac{9}{13}.\end{align}$

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Question

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

Answer

$\begin{align}&\text{The function }f(x)=200x - 320x^{2} + 250\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&-10\cdot (8x + 5)\cdot (4x - 5)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(200)\pm\sqrt{(200)^2-4(-320)(250)}}{2(-320)}=\frac{-200\pm600}{-640}\&\text{Either way, we find thatthe function crosses the x-axis at} x=\frac{5}{4}\text{ and }-\frac{5}{8}.\end{align}$

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Question

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

Answer

$\begin{align}&\text{The function }f(x)=732x - 96x^{2} - 702\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&-6\cdot (8x - 9)\cdot (2x - 13)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(732)\pm\sqrt{(732)^2-4(-96)(-702)}}{2(-96)}=\frac{-732\pm516}{-192}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{13}{2}\text{ and }\frac{9}{8}.\end{align}$

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Question

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

Answer

$\begin{align}&\text{The function }f(x)=524x + 272x^{2} + 84\text{ crosses the x-axis}\&\text{when it has a value of zero. To find the x values that}\&\text{satisfy this, we can either factor the equation:}\&4\cdot (4x + 7)\cdot (17x + 3)\&\text{or, if this is not intuitive, use the quadratic formula:}\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\&\frac{-(524)\pm\sqrt{(524)^2-4(272)(84)}}{2(272)}=\frac{-524\pm428}{544}\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{7}{4}\text{ and }-\frac{3}{17}.\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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.)

Answer

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.

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