All categories

3 years ago

5x³+8x²-7x-6

Mathematics

1 answer:

Lady_Fox [76]3 years ago

5 0

Answer:

Step-by-step explanation:

Whether we divide using long division or using synthetic division, the rule is the same: If, after division, there is no remainder (i. e., the remainder is zero), the divisor binomial is a factor or the associated root is indeed a root/zero/solution.

Divide 5x³+8x²-7x-6 by (x+2) using synthetic division. Use the divisor -2 (which comes from letting x+2 = 0):

--------------------------

-2 / 5 8 -7 -6

-10 4 6

------------------------------

5 -2 -3 0 Since the remainder here is 0, we know that

-2 is a root of 5x³+8x²-7x-6 and that (x+2) is

a factor of 5x³+8x²-7x-6.

Now check out the possibility that (x+1) is a factor of 5x^3 + 8x^2 - 7x - 6:

Use -1 as the divisor in synthetic division:

--------------------------

-1 / 5 8 -7 -6

-5 -3 10

------------------------------

5 3 -10 4

Since there is a non-zero remainder (4), we can conclude that (x + 1) is NOT a factor of the given polynomial expression.

You might be interested in

Find the missing side length

Mashcka [7]

It’s not 3 or 4 so I think 2

7 0

3 years ago

I'm kind of unsure of my answer, please help me. I appreciate your answers.

pentagon [3]

Answer:

We can call the numbers x and x + 1. Then we can write:

x + x + 1 = 113 which simplifies to 2x + 1 = 113.

5 0

3 years ago

Read 2 more answers

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

4 years ago

Geometry math question no Guessing and Please show work thank you

Snezhnost [94]

3m+21=6m
=>21=6m-3m
=>21=3m
=>7=m
Therefore option A is the answer

7 0

4 years ago

Read 2 more answers

I need to fill in the blanks for this equation and I cant figure it out

Alex17521 [72]

Answer:

1. +12x and 15

2. 14x

3. 2

Step-by-step explanation:

Step 1: Distribute -3

To distribute -3, you would multiply both -4x and -5 by -3, changing them to 12x and 15

Step 2: Combine 12x and 2x

Then you would combine like terms, meaning that you would add 12x and 2x, creating 14x.

Step 3: Combine 15 and -13

You would finish combining like terms by adding 15 to -13, creating 2

8 0

3 years ago