College Algebra - Finding Zeros of a Polynomial

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

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

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

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

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

Explanation

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.

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

x = 5, 2

x = –4, –3

x = –5, –2

x = 4, 3

x = 5

Explanation

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

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

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

$\small x=\pm11$

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

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

$\small x=0$

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

Find the roots of the following quadratic expression:

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

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

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

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

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

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.

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}.

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

$x=\frac{16}{7},-\frac{16}{25}$

$x=-\frac{16}{7},\frac{16}{25}$

$x=\frac{64}{7},-\frac{64}{25}$

$x=-\frac{50656}{175},-\frac{50144}{175}$

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

Find the zeros of the given polynomial:

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

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

Explanation

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

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

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

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

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

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

Consider the polynomial

$p(x) = x^{6}+ 6x^{4} -8x^{2}+ 1$

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

Hint: Think "Rational Zeroes Theorem".

The only rational zeroes of are and 1.

The only rational zero of is .

The only rational zero of is 1.

has at least one rational zero, but neither nor 1 is a zero.

has 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 $\left { -1, 1\right }$ .

Both values can be tested as follows:

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:

$p(x) = x^{6}+ 6x^{4} -8x^{2}+ 1$

1 is indeed 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. However, as their are no odd-degree coefficients, the sum is the same:

$p(x) = x^{6}+ 6x^{4} -8x^{2}+ 1$

is also a zero.

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

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

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

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

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

Explanation

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

Finding Zeros of a Polynomial

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation