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2 years ago

A(x + 1) - b(x + 1) - x - 1

Mathematics

1 answer:

Andrei [34K]2 years ago

8 0

Answer:

Step-by-step explanation:

Solution

I'm assuming you want this simplified. If not leave a note.

a(x + 1) - b(x + 1) - x - 1 remove the brackets

ax + a - bx - b - x - 1 gather like terms

ax - bx - x + a - b - 1

x(a - b - 1) + (a - b - 1) Use the distributive property to simplify

Answer

(x + 1)(a - b - 1)

The common factor is (a - b - 1). That can be pulled out on either side of the isolated + sign

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Factor completely: 2x4 − 32.

vazorg [7]

Answer:

Option b) is correct.

The completed factor of given expression is $2(x-2)(x+2)(x^2+4)$

Step-by-step explanation:

Given expression is $2x^4-32$

To find the completed factor for the given expression:

$2x^4-32$ :

Taking the common number "2" outside to the above expression we get

$2x^4-32=2(x^4-16)$

Now rewritting the above expression as below

$=2(x^4-2^4)$ (since 16 can be written as the number 2 to the power of 4)

$=2((x^2)^2-(2^2)^2)$

The above expression is of the form $a^2-b^2=(a+b)(a-b)$

Here $a=x^2$ and $b=2^2$

Therefore it becomes

$=2(x^2+2^2)(x^2-2^2)$

$=2(x^2+4)(x^2-2^2)$

The above expression is of the form $a^2-b^2=(a+b)(a-b)$

Here $a=x$ and $b=2$

Therefore it becomes

$=2(x^2+4)(x+2)(x-2)$

$=2(x+2)(x-2)(x^2+4)$

Therefore $=2(x+2)(x-2)(x^2+4)$

$2x^4-32=2(x-2)(x+2)(x^2+4)$

Option b) is correct.

The completed factor of given expression is $2(x-2)(x+2)(x^2+4)$

3 0

4 years ago

Read 2 more answers

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$