Question

plz tell the answer for 10th and 11 Qs

Answer 1

Hope it helps you.....

Answer 2

(10).

Given f(x) = 2x^4 + x^3 - 14x^2 - 19x - 6.

Given g(x) = x^2 + 3x + 2.

By splitting middle term, we get

⇒ x^2 + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 1)(x + 2).

Now,

If (x + 1) and (x + 2) are the factors of f(x), then f(x) is divisible by g(x).

If, f(-1) = 0 and f(-2) = 0, then f(x) is exactly divisible by g(x).

(i)

f(-1) = 2(-1)^4 + (-1)^3 - 14(-1)^2 - 19(-1) - 6

     = 2 - 1 - 14 + 19 - 6

     = 0.

(ii)

f(-2) = 2(-2)^4 + (-2)^3 - 14(-2)^2 - 19(-2) - 6

      = 32 - 8 - 56 + 38 - 6

      = 0.

Here, f(-1) = 0 and f(-2) = 0.

Therefore, f(x) is divisible by g(x).

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(11)

Given : (a + b + c)^3 - a^3 - b^3 - c^3

⇒ a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc - a^3 - b^3 - c^3

⇒ 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc

⇒ 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 + 2abc)

⇒ 3(a^2b + a^2c + abc + ac^2 + ab^2 + abc + b^2c + bc^2)

⇒ 3(a^2 + ac + ab + bc)(b + c)

⇒ 3(a(a + c) + b(a + c))(b + c)

⇒ 3(a + b)(a + c)(b + c).

Hope it helps!