Multiple polynomial the (x^3+3x^2+1)(3x^2+6x-2)

Mathematics

1 answer:

Mrrafil [7]3 years ago

6 0

The product of given polynomials is: $3x^5+15x^4-16x^3-3x^2+6x-2$

Step-by-step explanation:

Given polynomials are:

$(x^3+3x^2+1)\\and\\(3x^2+6x-2)$

In order to multiply both polynomials, we will use the distributive property

So,

$= (x^3+3x^2+1)(3x^2+6x-2)\\=x^3(3x^2+6x-2)+3x^2(3x^2+6x-2)+1(3x^2+6x-2)\\=3x^5+6x^4-2x^3+9x^4+18x^3-6x^2+3x^2+6x-2$

Combining like terms

$=3x^5+6x^4+9x^4-2x^3+18x^3+3x^2-6x^2+6x-2\\=3x^5+15x^4-16x^3-3x^2+6x-2$

Hence,

The product of given polynomials is: $3x^5+15x^4-16x^3-3x^2+6x-2$

Keywords: Polynomials, variables

Learn more about polynomials at:

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Answer is 100 as the next would be 200

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2 years ago

Find the area of the shaded region

o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

$\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)$

$A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}$