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Answer 1

Question - 1 :

(2x + 3) (6x + 3)

To find the degree of this, multiply it.

➙ (2x + 3) (6x + 3)

➙ 2x (6x + 3) + 3(6x + 3)

➙ 12x² + 6x + 18x + 9

➙ 12x² + 24x + 9

The degree of this polynomial is 2. ( It's a quadratic polynomial.)

Question - 2 :

$\sf P(x)=\dfrac{7x^{5/2}+2\sqrt{x^5}}{\sqrt{x^3}} \\\\ \sf P(x)=\dfrac{7x^{5/2}+2x^{5/2}}{x^{3/2}} \\\\ \sf P(x)=\dfrac{x^{5/2}(7+2)}{x^{3/2}} \\\\ \sf P(x)=x^{5/2-3/2} (7+2) \\\\ \sf P(x)=x^{2/2}(9) \\\\ \sf P(x)=9x$

Since the degree of the P(x) is 1, it's a monomial.

Used identities :

 $\boxed{\sf \sqrt{a^n}=a^{n/2}} \\\\ \boxed{\sf \dfrac{a^m}{a^n} =a^{m-n}}$

Question - 3 :

By long division method,

     $\Large \begin{array} {c|c|c} \sf x+2 & \sf x^3+0x^2+0x+18 & \sf x^2-2x+4 \\ & \sf x^3+2x^2 \qquad \qquad& \\ \cline{2-2} & \sf \qquad -2x^2+0x+18 & \\ & \sf -2x^2-4x & \\ \cline{2-2} & \sf \qquad \qquad 4x+18 & \\ & \sf \qquad \qquad 4x+8 \\ \cline{2-2} & \sf \qquad \qquad 10 \end{array}$

Remainder = 10

Question - 4 :

2x - 1 is a factor

⇒ 2x - 1 = 0

    2x = 1

    x = 1/2 = 0.5

It is a factor of 2x² + kx, so when we substitute x = 0.5 the result is zero.

2(0.5)² + k(0.5) = 0

  2(0.25) + 0.5k = 0

  0.5 + 0.5k = 0

  0.5k = -0.5

   k = -0.5/0.5

    k = -1

∴ The value of k is "-1"

Question - 5 :

x = -2,

5(-2) + 4(-2)² + 3(-2)³ + 5

-10 + 4(4) + 3(-8) + 5

-10 + 16 - 24 + 5

-13