Question

please give my answer ​

Answer 1

Step-by-step explanation:

(i)

${x}^{3} + x - 3 - 3 {x}^{2} = >x( {x}^{2} + 1) -$

$3(1 + {x}^{2}) = 0 = > (x - 3)(1 + {x}^{2}) =$

$0 = >$

So (1+ x square) can't be equal to 0 so x=3 hence there is only one root for this polynomial. (b) is the correct option.

(ii)

we need to substitute (2,-3) in

kx-3y+5=0 =>k(2)-3(-3)+5=0 =>2k+14=0 =>k= -7. So (b) is the correct option.

Hope it helps you.

Answer 2

Answer:

(i) 3

As the degree of the polynomial is the no. of zeros of the polynomial.

(ii)-7

kx-3y+5. (2,-3)

x=2, y=-3

Therefore,

k(2)-3(-3)+5=0

2k-(-9)+5=0

2k+9+5=0

2k+14=0

2k=-14

k=-7