Question

Maths
Factorisation
please factorise and help mr fast!!!!

Answer 1

Given Equation is (3a - 1)^2 - 6a + 2.

= > 9a^2 + 1 - 6a - 6a + 2

= > 9a^2 - 12a + 3

= > 3(3a^2 - 4a + 1)

= > 3(3a^2 - a - 3a + 1)

= > 3(a(3a - 1) -(3a - 1))

= > 3(a - 1)(3a - 1).

Hope this helps!

Answer 2

${(3a - 1)}^{2} - 6a + 2 \\ identity \: being \: used \: is: \\ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \\ \\ {(3a - 1)}^{2} \\ = {(3a)}^{2} - 2(3a)(1) + {1}^{2} \\ = 9 {a}^{2} - 6a + 1 \\ \\ (9 {a}^{2} - 6a + 1) - 6a + 2 \\ = 9 {a}^{2} - 12a + 3$
$9 {a}^{2} - 12a + 3 \\ sum = - 12 \\ product = 27 \\ \\ 9 {a}^{2} - 12a + 3 \\ = 9 {a}^{2} - 9a - 3a + 3 \\ = 9a(a - 1) - 3(a - 1) \\ = (9a - 3)(a - 1) \\ = 3(3a - 1)(a - 1)$
$therefore \: 3 \: \: and \: \: 3a - 1 \: and \: \: a - 1 \: are \: the \: factors \: of \: {(3a - 1)}^{2} - 6a + 2$