Question

If ( x-1 ) is the H.C.F. of (x²-1) and px²+q(x+1) then​

Answer 1

Answer:

Step-by-step explanation:

 x³ - px² + qx - 6

                                            x-1

                                            x = 1

1³ - p(1)² + q(1) - 6 = 0

1 -p +q -6 = 0

-p +q -6 +1 = 0

-p + q -5 = 0

p - q = -5 ---------- (i)

x-2

x=2

2³ - p(2)² + q(2) - 6 = 0

8 - 4p + 2q - 6 = 0

-4p +2q - 6 + 8 = 0

-4p + 2q + 2 = 0

4p -2q = 2 ---------- (ii)

p - q = -5 x4

4p -2q = 2 x1

4p -4q = -20 ----------- (iii)

4p - 2q =2 ------------ (iv)

Subtract equ iv from equ iii

-2q = -22

q = 11

p - q = -5

p - 11 = -5

p = -5 + 11

p = 6

Answer 2

If ( x-1 ) is the H.C.F. of (x²-1) and px² + q(x+1) then p + 2q = 0

Given : ( x-1 ) is the H.C.F. of (x²-1) and px² + q(x+1)

To find : The condition

Solution :

Here it is given that ( x-1 ) is the H.C.F. of (x²-1) and px² + q(x+1)

So px² + q(x+1) is completely divisible by x - 1

The zero of the polynomial x - 1 is given by

x - 1 = 0

⇒ x = 1

So by the given condition

$\displaystyle \sf{ \implies p. {1}^{2} + q(1 + 1) = 0 }$

$\displaystyle \sf{ \implies p + 2q= 0 }$

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