Question

Factorize: r^4 – 1



please answer it fast
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Answer 1

$\color{red} \large\bold{Given,} \\ \\ \sf \: \bull \: Length \: of \: the \: rectangle \: is \: 4 \: meters \: less \: than \: its \: breadth. \\ \\ \sf \bull \: Perimeter \: of \: the \: rectangle \: is \: 52 \: meters. \\ \\ \color{lime} \large \bold{To \: find \: out,} \\ \\ \sf \: \star \: Its \: length \: and \: breadth. \\ \\ \color{gold} \large\bold{Solution\: ―} \\ \\ \sf \: Let\:the\: \: breadth\:of\:the\:rectangle\:be\:x\:meters \\ \\ \sf Then,\:Length \:=\:2x\:-\:4 \\ \\ \color{orange}\boxed{ \sf \: Perimeter_{(rectangle)} \: = \:2 \: \times \: (Length \: \times \: B readth )} \\ \\ \sf \: \large \: Substituting\:the\:values,\:we\:get, \\ \\ \sf \: 52 = 2(2x - 4 + x) \\ \\ \sf \: 52 = 2(3x - 4) \\ \\ \sf \: 3x - 4 = \frac{52}{2} \\ \\ \sf \: 3x - 4 = 26 \\ \\ \sf \: 3x = 26 + 4 \\ \\ \sf \: x \: = \: \frac{30}{3} \\ \\ \sf \: x \: = \: 10 \\ \\ \sf \: Length \: = \: 2x - 4 \\ \\ \sf \: = \: 2 \times 10 - 4 \\ \\{ \underline{ \sf \: ☛ \: Length \: = \: 16 \: meters.}} \\ \\ \sf \: \:{ \underline{☛\: Breadth \: = \: 10 \: meters}}$

$\color{lime}\rule{200000000 pt}{2pt}$

$\color{red}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

We have,

${r}^{4} - 1$

$= \left({r}^{2} \right)^{2} - \left(1\right)^{2}$

$= \left({r}^{2} - 1\right)\left( {r}^{2} + 1\right)$

$= \left(r - 1\right)\left(r + 1\right)\left( {r}^{2} + 1\right)$