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College Algebra : Finding Zeros of a Polynomial

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Example Questions

Example Question #1452 : Algebra Ii

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

Possible Answers:

$\small x=-3\ or\ 4$

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

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

$\small x=-4\ or\ 3$

Correct answer:

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

Explanation:

Factor:

0=x^2+4x-12

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

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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Example Question #14 : How To Find The Solution To A Quadratic Equation

Solve for x.

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

Possible Answers:

x = 5

x = 4, 3

x = –4, –3

x = 5, 2

x = –5, –2

Correct answer:

x = 5, 2

Explanation:

1) 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!

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

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

x(x-2)-5(x-2)=0

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

(x-5)(x-2)=0

4) 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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Example Question #3 : Finding Roots

Solve for :

x^2+2 = 6x-6

Possible Answers:

x=1,-2

x=-2,-4

x=2,4

x=1, 2

Correct answer:

x=2,4

Explanation:

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:

x^2+2 = 6x-6

x^2-6x+2 = -6

x^2-6x+8 = 0

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$

(x-2)(x-4)=0

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

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

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Example Question #1 : Solving Quadratic Functions

Find the roots of the following quadratic expression:

6x^2 + 5x - 4

Possible Answers:

$x = \left \{ -4, 2\right \}$

$x = \left \{ -\frac{4}{3}, \frac{1}{2}\right \}$

$x = \left \{ -\frac{1}{2}, \frac{4}{3}\right \}$

$x = \left \{ -\frac{3}{4}, 2\right \}$

$x = \left \{ -1, 4\}$

Correct answer:

$x = \left \{ -\frac{4}{3}, \frac{1}{2}\right \}$

Explanation:

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.

6x^2 + 5x - 4 = 0

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.

6 * -4 = -24

-24 = 8 * -3

8 + -3 = 5

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

6x^2 + 5x - 4 = 0

6x^2 + 8x - 3x - 4 = 0

(6x^2 + 8x) + (- 3x - 4) = 0

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.

(3x(2x) + 4(2x)) + ((-1) 3x +(-1) 4) = 0

2x(3x+4)) + -1( 3x + 4) = 0

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

(3x+4)(2x-1)=0

Setting each factor = 0 we can find the solutions.

2x - 1 =0

2x = 1

x = 1/2

3x + 4 = 0

3x = -4

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

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

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Example Question #1 : Finding Zeros Of A Polynomial

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

Possible Answers:

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

$\small x=\pm11$

$\small x=\pm\frac{11}{9}$

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

$\small x=0$

Correct answer:

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

Explanation:

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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Example Question #2 : Finding Zeros Of A Polynomial

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

3+i

Possible Answers:

unknown

3+i

3-i

i-3

Correct answer:

3-i

Explanation:

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:

3-i

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Example Question #1 : Finding Zeros Of A Polynomial

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

Possible Answers:

$x=\frac{5}{4},-\frac{5}{8}$

$x=-\frac{3215}{16},-\frac{3185}{16}$

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

$x=-\frac{5}{4},\frac{5}{8}$

Correct answer:

$x=\frac{5}{4},-\frac{5}{8}$

Explanation:

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

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Example Question #2 : Finding Zeros Of A Polynomial

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

Possible Answers:

$x=-\frac{11755}{16},-\frac{11669}{16}$

$x=-\frac{13}{2},-\frac{9}{8}$

$x=26,\frac{9}{2}$

$x=\frac{13}{2},\frac{9}{8}$

Correct answer:

$x=\frac{13}{2},\frac{9}{8}$

Explanation:

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

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Example Question #3 : Finding Zeros Of A Polynomial

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

Possible Answers:

$x=\frac{7}{8},-\frac{9}{13}$

$x=-\frac{7}{2},\frac{36}{13}$

$x=-\frac{7741}{208},-\frac{8067}{208}$

$x=-\frac{7}{8},\frac{9}{13}$

Correct answer:

$x=-\frac{7}{8},\frac{9}{13}$

Explanation:

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

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Example Question #9 : Finding Zeros Of A Polynomial

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

Possible Answers:

$x=-7,-\frac{12}{17}$

$x=\frac{7}{4},\frac{3}{17}$

$x=-\frac{71157}{136},-\frac{71371}{136}$

$x=-\frac{7}{4},-\frac{3}{17}$

Correct answer:

$x=-\frac{7}{4},-\frac{3}{17}$

Explanation:

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

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College Algebra : Finding Zeros of a Polynomial

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1452 : Algebra Ii

Example Question #14 : How To Find The Solution To A Quadratic Equation

Example Question #3 : Finding Roots

Example Question #1 : Solving Quadratic Functions

Example Question #1 : Finding Zeros Of A Polynomial

Example Question #2 : Finding Zeros Of A Polynomial

Example Question #1 : Finding Zeros Of A Polynomial

Example Question #2 : Finding Zeros Of A Polynomial

Example Question #3 : Finding Zeros Of A Polynomial

Example Question #9 : Finding Zeros Of A Polynomial

All College Algebra Resources

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