Question

Factorise $a^{2}b+a^{2}c+a{b}^2+b^{2}c+a{c}^2+b{c}^2+3abc$.

• A. $(ab+ac+bc)(a+b+c)$
• B. $(ab-ac-bc)(a+b+c)$
• C. $(ab+ac-bc)(a-b+c)$
• D. $(ab+ac+bc)(a-b-c)$

Answer 1

The correct option is A $(ab+ac+bc)(a+b+c)$

We can take a look at this polynomial and found that the terms $(a^{2}b+a{b}^2)$ $\&$ $(a^{2}c+a{c}^2)$ $\&$ $(b{c}^2+b^{2}c)$ have $ab,ac\&bc$ in common respectively.

Therefore grouping the given polynomial as :

$\Rightarrow a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2+3abc$

$=ab(a+b)+ac(a+c)+bc(b+c)+3abc$

$=\{ab(a+b)+abc\}+\{ac(a+c)+abc\}+\{bc(b+c)+abc\}$

$=\left\{ab(a+b+c)\right\}+\left\{ac(a+b+c)\right\}+\left\{bc(a+b+c)\right\}$

$=(ab+bc+ac)(a+b+c)$

Hence, the answer is $(ab+bc+ac)(a+b+c).$