Determine the total number of roots of each polynomial function. f (x) = 3x^6 + 2x^5 + x^4 - 2x^3

Mathematics

1 answer:

iogann1982 [59]2 years ago

6 0

The total number of roots of the given polynomial function is 6.

The given polynomial is $f(x)=3x^{6}+2x^{5}+x^{4} -2x^{3}$ .

We need to determine the total number of roots of the given polynomial function.

What is the total number of roots of the polynomial function?

The number of roots of any polynomial is depended on the degree of that polynomial. Suppose n is the degree of a polynomial f(x), then fx) has n number of roots.

Here, the degree of the given polynomial function is 6.

Therefore, the given polynomial function will be having 6 roots.

To learn more about the polynomial function visit:

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in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$