Question
Factorise: $m^4- 256$
✓
Answer
We know that, $\mathrm{m}^4=\left(\mathrm{m}^2\right)^2$
and $256=(16)^2$
Therefore, $\mathrm{m}^4-256=\left(\mathrm{m}^2\right)^2-(16)^2$
$=\left(m^2+16\right)\left(m^2-16\right)[$using identity $\left.a^2-b^2=(a+b)(a-b)\right]$
$=\left(m^2+16\right)\left(m^2-4^2\right)$
$=\left(m^2+16\right)(m+4)(m-4)[$again, using identity $\left.a^2-b^2=(a+b)(a-b)\right]$
and $256=(16)^2$
Therefore, $\mathrm{m}^4-256=\left(\mathrm{m}^2\right)^2-(16)^2$
$=\left(m^2+16\right)\left(m^2-16\right)[$using identity $\left.a^2-b^2=(a+b)(a-b)\right]$
$=\left(m^2+16\right)\left(m^2-4^2\right)$
$=\left(m^2+16\right)(m+4)(m-4)[$again, using identity $\left.a^2-b^2=(a+b)(a-b)\right]$
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Name the property under multiplication used in each of the following
(i) $\frac{-4}{5} \times 1=1 \times\left(\frac{-4}{5}\right)=-\frac{4}{5}$
(ii) $\frac{-13}{17} \times\left(\frac{-2}{7}\right)=\frac{-2}{7} \times\left(\frac{-13}{17}\right)$
(iii) $\frac{-19}{29} \times\left(\frac{29}{-19}\right)=1$
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(iii) $\frac{-19}{29} \times\left(\frac{29}{-19}\right)=1$
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