Math / Algebra

QuestionQuestion 2 of 6, Step 3 of 4 4/15 Correct
Consider the following polynomial function. $f(x)=x^{4}-6 x^{3}+3 x^{2}+30 x-40$
Step 3 of 4 : Find the $x$ -intercept(s) at which $f$ crosses the axis. Express the intercept(s) as ordered pair(s).
Answer
Select the number of $x$ -intercept(s) at which $f$ crosses the axis.
Selecting an option will display any text boxes needed to complete your answer. none 1 2 3 4

Studdy Solution

STEP 1

What is this asking? We're looking for the points where this curvy polynomial graph crosses the x-axis, not just touches it! Watch out! Just because the polynomial has a degree of 4 doesn't mean it crosses the x-axis four times; it could cross fewer times or even not at all!
Also, remember that touching the x-axis isn't the same as crossing it.

STEP 2

1. Find all the x-intercepts.
2. Figure out which ones are crossings.

STEP 3

X-intercepts are where our graph crosses or touches the x-axis.
At these points, the y-value ( $f(x)$ ) is always zero.
So, we need to solve $f(x) = 0$ .

STEP 4

We're solving for $x$ in the equation: $x^{4} - 6x^{3} + 3x^{2} + 30x - 40 = 0$

STEP 5

From the previous steps (Step 1 and 2 of the original problem), we know that $x=2$ and $x=-2$ are roots.
This means $(x-2)$ and $(x+2)$ are factors.
So, we can factor those out.
Dividing the polynomial by $(x-2)(x+2) = x^2 - 4$ , we get $x^2 - 6x + 10$ .
So our factored polynomial is: $(x^2 - 4)(x^2 - 6x + 10) = 0$

STEP 6

$x^2 - 4 = 0$ gives us $x^2 = 4$ , so $x = 2$ and $x = -2$ .

STEP 7

$x^2 - 6x + 10 = 0$ .
This doesn't factor nicely, so let's use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a=1$ , $b=-6$ , and $c=10$ .
Plugging those in, we get: $x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$ $x = \frac{6 \pm \sqrt{36 - 40}}{2}$ $x = \frac{6 \pm \sqrt{-4}}{2}$ $x = \frac{6 \pm 2i}{2}$ $x = 3 \pm i$ These are imaginary roots!

STEP 8

We found four roots: $x = 2$ , $x = -2$ , $x = 3 + i$ , and $x = 3 - i$ .
Only real roots correspond to x-intercepts on the graph.
Our imaginary roots, $3 + i$ and $3 - i$ , don't show up as x-intercepts.

STEP 9

From the factored form $(x^2 - 4)(x^2 - 6x + 10)$ , we see that the $(x^2 - 4)$ factor gives us the roots $x=2$ and $x=-2$ .
Since each of these appears once (i.e., the factors are not raised to a higher power), the graph crosses the x-axis at these points.

STEP 10

The graph crosses the x-axis at two points: $(2, 0)$ and $(-2, 0)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find all the x-intercepts.
2. Figure out which ones are crossings.

STEP 3

X-intercepts are where our graph crosses or touches the x-axis.
At these points, the y-value ( $f(x)$ ) is always zero.
So, we need to solve $f(x) = 0$ .

STEP 4

We're solving for $x$ in the equation: $x^{4} - 6x^{3} + 3x^{2} + 30x - 40 = 0$

STEP 5

STEP 6

$x^2 - 4 = 0$ gives us $x^2 = 4$ , so $x = 2$ and $x = -2$ .

STEP 7

STEP 8

We found four roots: $x = 2$ , $x = -2$ , $x = 3 + i$ , and $x = 3 - i$ .
Only real roots correspond to x-intercepts on the graph.
Our imaginary roots, $3 + i$ and $3 - i$ , don't show up as x-intercepts.

STEP 9

From the factored form $(x^2 - 4)(x^2 - 6x + 10)$ , we see that the $(x^2 - 4)$ factor gives us the roots $x=2$ and $x=-2$ .
Since each of these appears once (i.e., the factors are not raised to a higher power), the graph crosses the x-axis at these points.

STEP 10

The graph crosses the x-axis at two points: $(2, 0)$ and $(-2, 0)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find all the x-intercepts.2. Figure out which ones are crossings.

STEP 3

X-intercepts are where our graph crosses or touches the x-axis.At these points, the y-value (f(x)f(x)f(x)) is always **zero**.So, we need to solve f(x)=0f(x) = 0f(x)=0.

STEP 4

We're solving for xxx in the equation: x4−6x3+3x2+30x−40=0x^{4} - 6x^{3} + 3x^{2} + 30x - 40 = 0x4−6x3+3x2+30x−40=0

STEP 5

STEP 6

x2−4=0x^2 - 4 = 0x2−4=0 gives us x2=4x^2 = 4x2=4, so x=2x = 2x=2 and x=−2x = -2x=−2.

STEP 7

STEP 8

We found four roots: x=2x = 2x=2, x=−2x = -2x=−2, x=3+ix = 3 + ix=3+i, and x=3−ix = 3 - ix=3−i.Only **real** roots correspond to x-intercepts on the graph.Our imaginary roots, 3+i3 + i3+i and 3−i3 - i3−i, don't show up as x-intercepts.

STEP 9

STEP 10

The graph crosses the x-axis at two points: (2,0)(2, 0)(2,0) and (−2,0)(-2, 0)(−2,0).

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1. Find all the x-intercepts.
2. Figure out which ones are crossings.

X-intercepts are where our graph crosses or touches the x-axis.
At these points, the y-value ( $f(x)$ ) is always zero.
So, we need to solve $f(x) = 0$ .

We're solving for $x$ in the equation: $x^{4} - 6x^{3} + 3x^{2} + 30x - 40 = 0$

$x^2 - 4 = 0$ gives us $x^2 = 4$ , so $x = 2$ and $x = -2$ .

We found four roots: $x = 2$ , $x = -2$ , $x = 3 + i$ , and $x = 3 - i$ .
Only real roots correspond to x-intercepts on the graph.
Our imaginary roots, $3 + i$ and $3 - i$ , don't show up as x-intercepts.

The graph crosses the x-axis at two points: $(2, 0)$ and $(-2, 0)$ .