Question

(3a-1)²-6a+2 factorise the term

Answer 1

$(3a-1)^{2} -6a+2 \\ \\ = (3a-1)(3a-1)-2(3a-1) \\ \\ =(3a-1-2)(3a-1) \\ \\ =(3a-3)(3a-1) \\ \\ =3(a-1)(3a-1)$

Answer 2

Answer:

The factorisation of $( 3 a - 1 ) ^ { 2 } - 6 a + 2$ will be $3(3a-1)(a-1)$

To find:

Factorise the given expression $( 3 a - 1 ) ^ { 2 } - 6 a + 2$

Solution:

Given,

Factorise the given expression,  

$( 3 a - 1 ) ^ { 2 } - 6 a + 2$

For factorising, splitting out the middle terms in the given expression, the middle term becomes as follows,

$\begin{array} { c } { ( 3 a - 1 ) ^ { 2 } - 6 a + 2 } \\\\ { = 9 a ^ { 2 } - 6 a + 1 - 6 a + 2 } \\\\ { = 9 a ^ { 2 } - 12 a + 3 } \end{array}$

As (-9) + (-3) = -12 and (-9) × (-3) = 27

$\begin{aligned} & = 9 a ^ { 2 } - 9 a - 3 a + 3 \\\\ = & 9 a ( a - 1 ) - 3 ( a - 1 ) \\\\ & = ( 9 a - 3 ) ( a - 1 ) \\\\ & = 3 ( 3 a - 1 ) ( a - 1 ) \end{aligned}$

Thus, the factorisation of the given expression $(3a-1)^2-6a+2$ will be $3(3a-1)(a-1)$