Question

12. If a, ß are the zeroes of the
polynomial x? – p(x +1) – C, then
(a + 1)(ß +1) = *
+​

Answer 1

Answer:

1 - c

Step-by-step explanation:

→ x² - p(x + 1) - c → x² - px - p - c

→ x² - px - (p + c)

Polynomials written in form of x² - Sx + P, represent S as sum of their roots and P as product of roots.

So, here, if a and ß are roots:

• a + ß = p

• aß = - (p + c)

So,

= > (a + 1)(ß + 1)

= > aß + a + ß + 1

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

= > - p - c + p + 1

= > 1 - c

Answer 2

Answer:

⭐ Given ⭐

✍ If a, ß are the zeroes of polynomial : x²-p(x+1)-c

⭐ To find ⭐

✍ (a + 1)(ß +1)

⭐ Solution ⭐

=> x²-p(x+1)-c

=> x²-px-p-c

=> x²-px-(p+c)

➡ Polynomial can be written as

✍ k(x² - SX + P)

Where,

⚫ S = Sum of the roots.

⚫ P = Product of the roots.

⭕ So here, α and β are the roots,

✏ α + β = p

✏ α × β = -(p+c)

➡ Now,

=> (a + 1)(ß +1)

=> aß + α + β + 1

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

=> 1 - c✔

✍ Hence, the value of

(a + 1)(ß +1) = 1 - c

Step-by-step explanation:

HOPE IT HELP YOU