Question

$\text{Factorisation of}a{x}^{2}y+bx{y}^2+cxyz\text{gives:}$

• A. $xy(ax+by+cz)$
• B. $x^{2}y(ax+by+cz)$
• C. $x(ax+b{y}^2+cz)$
• D. $xy(ax+by+cxz)$

Answer 1

The correct option is A $xy(ax+by+cz)$

The irreducible factor form of each of the terms is,
$a{x}^{2}y=a\times x\times x\times ybx{y}^2=b\times x\times y\times ycxyz=c\times x\times y\times z\text{The common factors are}x\text{and}y.\therefore a{x}^{2}y+bx{y}^2+cxyz=(a\times x\times x\times y)+(b\times x\times y\times y)+(c\times x\times y\times z)=(x\times y)\left[\right(a\times x)+(b\times y)+(c\times z\left)\right]=xy(ax+by+cz)$