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3 years ago

1. Which of the following in the process of finding the factors of the

Mathematics

1 answer:

Sidana [21]3 years ago

6 0

Answer:

We cannot say it's different Difference ot of Two Cubes because 2d2 is not not cube it's square. and 8d is not a cube.

We cannot say Difference of Two Squares because only first term 2d2 has a square.

It is not a Perfect Square Trinomials because Perfect Square Trinomials appears as ax2 + bx + c, but the given ter doesn't follow this.

A. Common Monomial Factor can be regarded as a variable, or more than one variable that that is present in the polynomial terms.

Example is 4x2 + 16x,

If factorize we have 4x(x + 4) with monomer of 4x and polynomial of x + 4). that cannot be factorized into lower polynomial.

Hence 2d2-8d can be factor as 2d(d-4) where 2d is the monomer and (d-4) cannot be factorize into lower degree polynomial.

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3 years ago

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What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

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1 year ago