Question

Factorization : {3×2=6}

(1) 12a to the power of 2 b + 15ab to the power of 2


solve this plz ​

Answer 1

Factorise:

• 12a²b + 15ab²

Factors of 12a²b = 2 × 2 × 3 × a × a × b

Factors of 15ab² = 3 × 5 × a × b × b

Common factors = 3 × a × b

Therefore, 3ab

→ 12a²b + 15ab²

→ (2 × 2 × 3 × a × a × b) + (3 × 5 × a × b × b)

→ 3ab(4a + 5b)

Henceforth, solved!

Additional information:

Some factorising identities!

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$