Question

$factorise 16a {}^{2} + 9b {}^{2} + 4c {}^{2} + 24ab + 12bc + 16ca$
​

Answer 1

If we carefully observe ; the given problem is in the form of :

$\sf{{(a + b + c)}^{2} = {(a)}^{2} + {(b)}^{2} + {(c)}^{2} + 2ab + 2bc + 2ca}$

So factorization will also be on the same basis :

$\longrightarrow{\sf{{16a}^{2}\: can \: be \: written \: as \implies {(4a)}^{2}}}$

$\longrightarrow{\sf{{9b}^{2}\: can \: be \: written \: as \implies {(3b)}^{2} }}$

$\longrightarrow{\sf{{4c}^{2} \:can \: be \: written \: as \implies {(2c)}^{2} }}$

Taking up the simplified values of a , b and c from above simplifications ; we can simplify the other parts:

$\sf{ (24ab)\: can \: be \: written \: as \implies 2 \times 4a \times 3b }$

$\sf{ (12bc)\: can \: be \: written \: as \implies 2 \times 3b \times 2c}$

$\sf{ (16ca)\: can \: be \: written \: as \implies 2 \times 4a \times 2c }$

So ; the factors are :

$\sf{(4a + 3b + 2c) and (4a + 3b + 2c)}$

(or)

$\sf{{(4a + 3b + 2c)}^{2}}$