Question

1. Zero of the polynomial p(x) = (cx + d) is

a) –d b) -c c)d/c d) –d/c

2. If x = 2 is a zero of the polynomial 2x^2 + 3x – p, then the value of p is
a) -4 b) 0 c) 6 d) 14
3. If one zero of the polynomial 3x^2 – 10x + p is 1/3, then the value of p and other zero respectively is
a) 0 b) 5 c) 1/6 d) 6

4. If p(x) = x^2 - 2√2x + , then p(2√2) is equal to

a) 0 b) 1 c) 4√2 d) 8√2 + 1

Please Answer All 4 questions in 1 post
And I will mark the best answer as brainliest
and one more things these questions are from the chapter polynomials....
Please Help

Answer 1

1) d ( Solve cx+d =0)
2) d (Put x=2 in p(x)=0)
3) All options are incorrect. Correct ans is 3
4) a ( Put this value in place of x)

Answer 2

Step-by-step explanation:

1)a)Given p(x)=cx+d,c is not equal to 0,c,d are real numbers.

To find zero of a polynomial we put p(x)=0 and find x

⇒cx+d=0

⇒cx=−d

⇒x= − d/c

Therefore x=−

is the zero of the given polynomial.

2. put the value of x = 2

(2×(2)2-3×2+7k = 0 then 8-6+7k=O

k= - 2/7

3. attached image

4. Given, p(x) = x2 – 2√2x + 1 …(i)

On putting x = 2√2 in Eq. (i), we get

P(2√2) = (2√2)2– (2√2)(2√2) + 1 = 8 - 8 + 1 = 1