Math, asked by suhas2006, 6 months ago

the largest number by which the expression x³+6x²+11x+6 is divisible for all possible integral value of X is​

Answers

Answered by jakharkapil59
1

Answer:

x = -1, -2, -3

Step-by-step explanation:

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)

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

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

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

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Answered by hukam0685
1

Step-by-step explanation:

Given that:

 {x}^{3}  + 6 {x}^{2}  + 11x + 6

To find:the largest number by which the expression is divisible for all possible integral value of X is

Solution:

The integral values by which the given polynomial is completely divisible are the zeros of the polynomial.

Thus,find the zeros of the polynomial.

In cubic polynomial first factor can be easily find by hit and trial

Put x=-1

p(x) =  {x}^{3}  + 6 {x}^{2}  + 11x + 6 \\  \\ p( - 1) =  {( - 1)}^{3}  + 6 {( - 1)}^{2}  + 11( - 1) + 6 \\  \\ =  - 1 + 6 - 11 + 6 \\  \\  =  - 12 + 12 \\  \\  p( - 1)= 0 \\  \\

Thus one zero is -1.

Now factor is (x+1).

For others two factors and zeros divide the polynomial by (x+1)

x + 1) {x}^{3}  + 6 {x}^{2}  + 11x + 6( {x}^{2} + 5x + 6  \\  \:  \:  \:  \: \:  \:  \:  \:  \:  \:   \:  {x}^{3}  +  {x}^{2}  \\ \:  \:  \: \:  \:  ( - ) \:  \:  \:  \:  \:  \: ( - )  \\  -  -  -  -  -  -  -  -  \\  \:  \:  \:  \:  \:  \:  \:  \: 5 {x}^{2}  + 11x \\  \:  \:  \:  \:  \:  \:  \:  \:  \: 5 {x}^{2}  + 5x \\  \:  \:  \:  \:  \:  \: ( - ) \:  \:  \:  \:  \: ( - ) \\  \:  \:  \:  \:   -  -  -  -  -  -  -  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 6x + 6 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 6x + 6 \\  \:  \:  \:  \:  \:  \:  \:  ( - ) \:  \:  \: ( - ) \\  -  -  -  -  -  -  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  0 \\

Now,factorise the quotient polynomial

 {x}^{2}  + 5x + 6 = 0 \\ \\   {x}^{2} + 3x + 2x + 6 =0  \\  \\ x(x + 3) + 2(x + 3) = 0 \\  \\ (x + 2)(x + 3) = 0 \\  \\ (x + 2) = 0 \\  \\ x =  - 2 \\  \\ or \\  \\ (x + 3) = 0 \\  \\ x =  - 3 \\  \\

The only values of x which can divide the polynomial are -1,-2 and -3.

There are only three integral values of x by which given polynomial is completely divisible,out of which -1 is largest.

Hope it helps you.

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