Question

How many roots are there in the polynomial equation x³+6x²+11x-6=0? Show complete solution
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Answer 1

Answer:

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Let p(x) = x 3 + 6x 2 + 11x + 6

Put x = – 1

p(– 1) = (– 1)3 + 6(– 1)2 + 11(– 1) + 6 = – 1 + 6 – 11 + 6 = 0

∴ (x + 1) is a factor of p(x)

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So, we cam break up terms of p(x) as follows.

p(x) = x 3 + 6x 2 + 11x + 6

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Therefore, the integral roots of the given equation is find out as:

p(x) = 0

⇒ (x+1)(x+2)(x+3) = 0

⇒ x = -1, -2, -3

hope this will help you~~~~

Answer 2

Answer:

⋆ ANSWER ⋆

(1)³ -6(1)² +11(1) -6=0

one root is (x-1)

divide whole eqn by (x-1)

(x-1) | x³ -6x² + 11x -6 | X²-5x +6

x³ -x²

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-5x²+11x-6

-5x²+5x

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6x-6

6x-6

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0

so eqn is x²-5x+6=0

x²-2x-3x+6=0

(x-2)(x-3)=0

so the roots are 1 ,2,3

upper user put "+6" instead of "-6" ( by mistake) u can change it ur own :)

hope this will help you~~~