Question

please solve these equation
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Answer 1

Answer:

(i). (x+2)(x+3)

(ii). (3x+1)(3x+1)

Step-by-step explanation:

(i) x²+5x+6

=>x²+3x+2x+6

=>x(x+3)+2(x+3)

=>(x+2)(x+3)

(ii) 9x²+6x+1

=>9x²+3x+3x+1

=>3x(3x+1)+1(3x+1)

=>(3x+1)(3x+1)

Answer 2

Step-by-step explanation:

1.

$x ^{2} + 2x +3x + 6 = 0 \\ x(x + 2) + 3(x+ 2) = 0 \\ (x + 2)(x + 3) = 0 \\ either \: x + 2 = 0 \\ x = - 2 \\ or \\ x + 3 = 0 \\ x = - 3$

2.

$9x^2 - 6x + 1 = 0 \\ 9x^{2} - 3x - 3x + 1 = 0 \\ 3x(3x - 1) - 1(3x - 1) = 0 \\ (3x - 1)(3x - 1) = 0 \\ therefore \: x = 1 \div 3$

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