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4 years ago

Factor by grouping 2u^3+3u^2-14u-21

Mathematics

2 answers:

nika2105 [10]4 years ago

8 0

(2u³+3u²)-(14u+21)
u²(2u+3)-7(2u+3)
(u²-7)(2u+3)
(u+√7)(u-√7)(2u+3)

mariarad [96]4 years ago

6 0

$\bf 2u^3+3u^2-14u-21\implies (2u^3+3u^2)-(14u+21)
\\\\\\
u^2(2u+3)-7(2u+3)\implies \stackrel{\stackrel{common}{factor}}{(2u+3)}(u^2-7)$

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3 years ago

Complete the table by converting each decimal to a fraction.

AlekseyPX

if you've read those links already, you'd know what we're doing here.

we'll move the repeating part to the left-side of the dot, by multiplying by "1" and as many zeros as needed, or 10 at some power pretty much.

on 0.13 we need 100 to get 13.13.... and on 0.1234, we need 10000 to get 1234.1234....

$\bf 0.\overline{13}~\hspace{10em}x=0.\overline{13}
\\\\[-0.35em]
~\dotfill\\\\
\begin{array}{|lll|ll}
\cline{1-3}
&&\\
100\cdot 0.\overline{13}& = & 13.\overline{13}\\
100\cdot x&& 13 + 0.\overline{13}\\
100x&&13+x
\\&&\\
\cline{1-3}
\end{array}\implies \begin{array}{llll}
100x=13+x\implies 99x=13
\\\\
x=\cfrac{13}{99}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
0.\overline{1234}~\hspace{10em}x=0.\overline{1234}
\\\\[-0.35em]
~\dotfill$

$\bf \begin{array}{|ccc|ll}
\cline{1-3}
&&\\
10000\cdot 0.\overline{1234}&=&1234.\overline{1234}\\
10000\cdot x&=&1234+0.\overline{1234}\\
10000x&=&1234+x
\\&&\\
\cline{1-3}
\end{array}\implies \begin{array}{llll}
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\\\\
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\end{array}$

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3 years ago

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