Question

(a²-a)(4a²-4a-5)-6 factorise please​

Answer 1

Step-by-step explanation:

$( {a}^{2} - a)( {4a}^{2} - 4a - 5) - 6$

$= {a}^{2} ( {4a}^{2} - 4a - 5) - a( {4a}^{2} - 4a - 5) - 6 = 0$

$= {4a}^{4} - {4a}^{3} - 5 {a}^{2} + {5a}^{3} + {4a}^{2} + 5a - 6 = 0$

$= {4a}^{4} + {a}^{3} - {a}^{2} + 5a - 6 = 0$

${a}^{2} ( {4a}^{2} + a - 1) + 5a - 6 = 0$

$we \: get \: two \: equation \: now$

${a}^{2} + 5a - 6 = 0 ....(1)\: \: \: \: \: \: \: and \: \: \: \: \: {4a}^{2} + a - 1 = 0.....(2)$

$\: \: \ {a}^{2} + 6a - a - 6 = 0 \: \: \: \: \: \: and \: {4a}^{2} + a - 1 = 0$

$\: \: \: \bf{ {a}^{2} + 6a - a - 6 = 0}$

$\: \: \bf{a(a + 6) - 1(a + 6) = 0}$

$\: \bf{(a + 6) = 0 \: \: \: and \: (a - 1) = 0}$

$\: \: \: \implies \bf{ \: a = - 6 \: \: and \: a = 1}$

$\: \: \: \bf{ {4a}^{2} + a - 1 = 0}.......(2)$

$\: \: \to\bf{x = \frac{ -b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\: \: \: \: \to \bf{x = \frac{ - 1 \pm \sqrt{ {1}^{2} - 4( - 1)} }{2 \times 4} }$

$\: \: \: \to \bf{x = \frac{ - 1 \pm \sqrt{5} }{8} }$

$\huge{\blue{\red{Answer}}}$

$\: \: \: \: \: \boxed { \bf \pink{x = \frac{ - 1 + \sqrt{5} }{8} \: \: and \: x = \frac{ - 1 - \sqrt{5} }{8} }}$

$\: \: \: \: \boxed { \bf \red{x = - 6 \: \: and \: x = 1}}$