Factorise:

Question:

Factorise:

(i) 24x3 − 36x2y

(ii) 10x3 − 15x2

(iii) 36x3y − 60x2y3z

Solution:

(i) H.C.F. of $24 x^{3}$ and $36 x^{2} y$ is $12 x^{2}$.

$\therefore 24 x^{3}-36 x^{2} y=12 x^{2}(2 x-3 y)$

(ii) H.C.F. of $10 x^{3}$ and $15 x^{2}$ is $5 x^{3}$

$\therefore 10 x^{3}-15 x^{2}=5 x^{2}(2 x-3)$

(iii) H.C.F. of $36 x^{3} y$ and $60 x^{2} y^{3} z$ is $12 x^{2} y$.

$\therefore 36 x^{3} y-60 x^{2} y^{3} z=12 x^{2} y\left(3 x-5 y^{2} z\right)$

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