Question

Factorise
a) x2-3x/2
b) 3x2-5x
c) 4x-2x2
d) 9x2-3x
e) 4p3-q3/2

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Answer 1

Answer:

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Answer 2

$\bf \huge{x}^{2} - \frac{3x}{2} \\ \\ \sf= > \red{ \: \: \: \: x(x - \frac{3}{2} \bigg)}$

$\bf \huge {3x}^{2} - 5x \\ \\ \sf = > \red{x(3x - 5)}$

$\bf \huge4x - {2x}^{2} \\ \\ \sf = > {2x(2 - {x}^{2}) } \\ \sf = > 2x[ {( \sqrt{2} )}^{2} - {(x)}^{2} ] \\ \sf= > 2x[( \sqrt{2} - x)( \sqrt{2} + x)] \\ \sf = > \red{2x( \sqrt{2} - x)( \sqrt{2} + x )}$

$\bf \huge {9x}^{2} - 3x \\ \\ \sf= > \: \: \: \red{ 3x(3x - 1)}$

$\bf \huge {4p}^{3} - \frac{ {q}^{3} }{2} \\ \\ \sf = > \bigg(\sqrt[3]{4}p \bigg)^{3} - \bigg( \frac{q}{ \sqrt[3]{2} } \bigg)^{3} \\ \\ \small{\sf = > \bigg( \sqrt[3]{4}p - \frac{q}{ \sqrt[3]{2} } \bigg ) \bigg [( \sqrt[3]{4}p)^{2} + \bigg( \frac{q}{ \sqrt[3]{2} } \bigg) ^{2} + \frac{ \sqrt[3]{4} \: pq }{ \sqrt[3]{2} } \bigg] }\\ \\ \small{ \sf = > \red{\bigg( \sqrt[3]{4}p - \frac{q}{ \sqrt[3]{2} } \bigg ) \bigg( \sqrt[3]{16} {p}^{2} + \frac{ {q}^{2} }{ \sqrt[3]{4} } + \frac{ \sqrt[3]{4} }{ \sqrt[3]{2} }pq \bigg)}}$

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