Question

factorise x cube + 6 X square + 11 x + 6

Answer 1

HEY FRIEND,
HERE IS YOUR ANSWER
${x}^{3} + 6 {x}^{2} + 11x + 6 \\ = {x}^{3} + {x}^{2} + 5 {x}^{2} + 5x + 6x + 6 \\ = {x}^{2} (x + 1) + 5x(x + 1) + 6(x + 1) \\ = (x + 1)( {x}^{2} + 5x + 6) \\ = (x + 1)(x + 2)(x + 3)$
I HOPE IT HELPS YOU...!!!

Answer 2

Answer: x³ + 6x² + 11x + 6 = (x + 1)(x + 2)(x + 3)

Step-by-step explanation:

It became easier because I was familiar with this polynomial!!! ;-)

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TO REMEMBER...

 

In a cubic polynomial ax³ + bx² + cx + d = 0, if the roots are α, β and γ, then,

$\alpha+\beta+\gamma=-\frac{b}{a} \\ \\ \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a} \\ \\ \alpha\beta\gamma=-\frac{d}{a}$

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Here, p(x) = x³ + 6x² = 11x + 6 = 0

a = 1

b = 6

c = -11

d = 6

Let the roots be α, β and γ.

Here it seems that,

$\alpha+\beta+\gamma=-\frac{6}{1}=-6=-1-2-3=(-1)+(-2)+(-3) \\ \\ \\ \alpha\beta+\beta\gamma+\gamma\alpha=\frac{11}{1}=11=2+6+3=(-1\times -2)+(-2 \times -3)+(-3 \times -1) \\ \\ \\ \alpha\beta\gamma=-\frac{6}{1}=-6=-1 \times -2 \times -3$

So the roots are,

$\alpha=-1 \\ \\ \beta=-2 \\ \\ \gamma=-3$

$\therefore\ x^3+6x^2+11x+6=(x+1)(x+2)(x+3)$

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Hope this may be helpful.

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Thank you. Have a nice day. :-))

   

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