Question

if A and B are the zeroes of the polynomial x^2-p(x+1)-c, then the value of (A+1) (B+1) is​

Answer 1

Step-by-step explanation:

Given :-

A and B are the zeroes of the polynomial x^2-p(x+1)-c.

To find :-

Find the value of (A+1) (B+1) ?

Solution :-

Given quardratic polynomial is x^2-p(x+1)-c.

Let P(x) = x^2-p(x+1)-c.

=> P(x) = x^2-px-p-c

=> P(x) = x^2-px-(p+c)

On comparing with the standard quadratic polynomial ax^2+bx+c,

We have ,

a = 1

b = -p

c = -(p+c)

Given that

The zeroes of P(x) are A and B

We know that

Sum of the zeroes = -b/a

=> A + B = -(-p)/1

=> A + B = p/1

A+B = p -------------(1)

Product of the zeroes = c/a

=> A × B = -(p+c)/1

=> A × B = -(p+c)

AB = -(p+c)---------(2)

Now

The value of (A+1) (B+1)

=> A (B+1) +1 (B+1)

=> AB + A + B + 1

=> (AB)+(A+B)+1

From (1)&(2)

=> -(p+c)+p+1

=>-p-c+p+1

=> (-p+p)-c+1

=> 0-c+1

=> -c+1

=> 1-c

Answer:-

The value of (A+1) (B+1) for the given problem is (1-c)

Used formulae:-

• The standard quadratic polynomial is ax^2+bx+c

• Sum of the zeroes = -b/a

• Product of the zeroes = c/a