Math, asked by aryansingh2970, 22 days ago

factorise (x³+6x²+11x+6)

Answers

Answered by ItzHannu001
0

Answer:

Let p(x) = (x³+6x²+11x+6), constant term of p(x) is 6

Factor of 6 is +1, +2, +3 and +6

Now, p(-1) = (-1)^3 + 6×(-1)^2 + 11×(-1) + 6

= -1 + 6×1 + (-11) +6

= -1 +6 -11 +6

= 0

By trial we find p(-1) =0 , so (x+1) is a factor of p(x)

 \tt  \implies{x}^{3}  + 6 {x}^{2}  + 11x + 6 =  {x}^{3}  +  {x}^{2}  +  {5x}^{2}  + 5x + 6x + 6 \\  \tt  =  {x}^{2} (x + 1) + 5x(x + 1) + 6(x + 1) \\  \tt = (x + 1)( {x}^{2}  + 5x + 6) \\  \tt = (x + 1)( {x}^{2}  + x - 6x + 6) \\  \tt = (x + 1){ {x(x + 1) - 6(x + 1)} \brack} \\  \tt = (x + 1)(x + 1)(x - 6)

Swipe right

Similar questions