Question

if (x + a)(x + b)(x + c) = x³-6x²+11x-6 , find the value of a²+b²+c²:

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Answer 1

Answer:

14

Step-by-step explanation:

Given :

if (x + a)(x + b)(x + c) = x³- 6x² + 11x - 6 , find the value of a²+b²+c²

Solution :

By factorizing the expression,

We get,

⇒ x³- 6x² + 11x - 6

⇒ x³- x² - 5x² + 5x + 6x - 6

⇒ x² (x - 1) - 5x (x - 1) + 6(x  - 1)

⇒ ( x² - 5x + 6 ) ( x - 1 )

⇒ (x² - 2x - 3x - 6) (x - 1)

⇒ (x (x - 2) - 3( x - 2) ) ( x - 1 )

⇒ (x - 3) (x - 2) (x - 1)

_

⇒ (x - 3) (x - 2) (x - 1) = (x + a)(x + b)(x + c)

By comparing,

⇒ a = -3 , b = -2 , c = -1

Hence,

⇒ a² + b² + c² = (-3)² + (-2)² + (-1)² = 9 + 4 + 1 = 14

Answer 2

Answer:

14

Step-by-step explanation:

Putting x =1

(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

_____________

-5x² +11x-6

-5x²+5x

_____________

6x-6

6x-6

_______

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

a=-1,b=-2,c=-3

a²+b²+c²=14

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