Question

factorise 9(x+2)^2-6(x+2)+1 the answer of the question is (3x+5)(3x+5) plz solve it.​

Answer 1

Question:-

• Factorise $9(x+2)^{2} - 6(x+2)+1$

Solution:-

Let $x+2=y$

Therefore,

$9(x + 2)^{2} - 6(x + 2) + 1$

$= 9 {y}^{2} - 6y + 1$

$= {(3y)}^{2} - 2 \times 3y \times 1 + {(1)}^{2}$

$= (3y - 1)^{2}$

Putting the value of $y$, we get,

$= (3(x + 2) - 1)^{2}$

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

$= {(3x + 5)}^{2}$

$= (3x + 5)(3x + 5)$

Answer:-

• $(3x + 5)(3x + 5)$

Answer 2

$\bf\large\blue{Question :-}$

$9 {(x + 2)}^{2} - 6(x + 2) + 1$

$\bf\large\blue{Solution :-}$

$\bf{Let's\:take\:(x + 2)\:as\:y.}$

$\bf{Therefore,}$

$9 {(x + 2)}^{2} - 6(x + 2) + 1$

$= \: 9 {y}^{2} - 6y + 1$

$\bf{Now, \:by\: inserting\: formula\: of \:factorisation \::-}$

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

$\bf{We\:get\::-}$

$(3 {y})^{2} - 2 \times 3y \times 1 + {(1)}^{2}$

$= (3y - 1)(3y - 1)$

$\bf{Now,\: putting\: value\: of \: y\::-}$

$[ {3(x + 2) - 1} ][ {3(x + 2) - 1} ]$

$= \: (3x + 5)(3x + 5)$

$\bf\large\blue{Answer :-}$

$(3x + 5)(3x + 5)$