Question

please solve the question, class 8, subject math, lesson factorisation, no wrong answer, if you spam I will report​

Answer 1

:-

${\huge{ \blue{ \overbrace{ \underbrace{\red{\mathcal{ \: @Πswēr:-}}}}}}}$

$6) \: \: \: \: {x}^{2} + 9x + 20 \\ = {x}^{2} + 4x + 5x + 20 \\ = x(x + 4) + 5(x + 4) \\ = (x + 4) \: \:( x + 5) \: answer$

$7) \: \: \frac{16 {m}^{2} + 40 mn + 25 {n}^{2}}{5 n+ 4m} \\ = \frac{{(4m) }^{2} + 2 \times 4m \times 5n + {(5n)}^{2} }{5 n+ 4m} \\ = \frac{ {(4m +5n)}^{2} }{4m + 5n} \\ = \frac{(4m + 5n)(4m + 5n)}{(4m + 5n)} \\ now \: (4m + 5n) \: gets \: cancelled \: \\ = (4m + 5n) \: \: \: answer$

$8) \: \: \: \frac{a²+10a-24}{a + 12 } \\ = \frac{{a}^{2} + 12a - 2a - 24}{a + 12} \\ = \frac{a(a + 12) \: - 2(a + 12)}{(a + 12)} \\ = \frac{(a - 2) \: (a + 12)}{a + 12} \\ now \: (a + 12) \: gets \: cancelled \\ = (a - 2) \: \: \: \: \: answer$

$9) \: \: {m}^{2} + 12m + 27 \\ = {m}^{2} + 3m + 9m+ 27 \\ = m(m + 3) \: + 9(m + 3) \\ = (m + 3) \: (m + 9) \: \: answer$

$10) \: \: \: \: 121 {m}^{2} - 64 {n}^{2} \\ = we \: know \: that, \\ = [(a²-b²)=(a-b)(a+b)] \\ = {(11m)}^{2} - {(8n)}^{2} \\ = (11m + 8n) \: \: (11m - 8n)$

${ \huge{ \red{ \boxed{ \ hope \: u \: understood \:\: }}}}$