Question

If (x+2) (x+5) is the HCF of the polynomials (x+2)(x^2+6x+a) and (x+5)(x^2+8x+b) then find the values of a and b

Answer 1

Answer:

a=5 and b=12

Step-by-step explanation:

HCF = (x+2)(x+5)

(x+2)(x+5) should be factor of (x+2)(x2+6x+a) hence by reminder theorem (x+5) can be factor of (x2+6x+a) only if x=-5.

=> (-5)2 + (6 x -5) + a = 0

=> 25 – 30 + a = 0

=> -5 + a = 0

=> a = 5.

 

(x+2)(x+5) should be factor of (x+5)(x2+8x+b) hence by reminder theorem (x+2) can be factor of (x2+8x+b) only if x=-2.

=> (-2)2 + (8 x -2) + b = 0

=> 4 – 16 + b = 0

=> -12 + b = 0

=> b = 12.

Therefore a=5 and b=12

Answer 2

Answer:

The value of a is 5 and b is 12.

Step-by-step explanation:

Given (x+2) (x+5) is the HCF of the polynomials (x+2)(x^2+6x+a) and (x+5)(x^2+8x+b).

Here two polynomials are ${x}^{2} + 6x + a \: \: and \: \: (x + 5)( {x}^{2} + 8x + b)$

It is confirm that (x+5) is the root of${x}^{2} + 6x + a \: \: = 0$

and (x+2) is the root of $(x + 5)( {x}^{2} + 8x + b) = 0$

Putting x=-5 in the polynomial (1),

${ (- 5)}^{2} + 6 \times ( - 5) + a = 0$

$25 - 30 + a = 0 \\ a = 5$

and putting x=-2 in the polynomial (2),

$( - 2 + 5)( {( - 2)}^{2} - 2 \times 8 + b) = 0 \\ 3 \times (b - 12) = 0 \\ b = 12$

So, value of a is 5 and b is 12.