Question

please if anyone from class 9 cbse please solve an send me 2nd chapter 2.2 exercise with solutions​

Answer 1

Note :

Make sure to post one question at a time in order to get a satisfying solution.

$\tt \: question$

Find the value of polynomial

$5x - {4x}^{2} + 3 \: at$

1. x=0
2. x= -1
3. x= 2

$\tt \: solution$

$p(x) = 5x - {4x}^{2} + 3$

Finding at p(x) = 0

*replacing the x with 0

$p(0) = 5 \times 0 - 4 \times {0}^{2} + 3 \\ \\ p(0) = 3$

Finding at p(x) = -1

*replacing the x with -1

$p( - 1) = 5 \times ( - 1) - 4 \times ( - {1)}^{2} + 3$

$\implies \: p( - 1) = - 5 - 4 + 3$

$p( - 1) = - 6$

Finding at p(x) = 2

*replacing the x with 2

$p(2) = 5 \times 2 - 4 \times {2}^{2} + 3$

$p(2) = 10 - 16 + 3 \\ = - 3$

$\tt \: question$

Find p(0) , p(1) and p(2) for each of the following.

p(y) = y^2 - y + 1

p(t) = 2+t+2t^2-t^3

p(x) = x^3

p(x) = (x-1) (x+1)

$\tt \: solution$

$p(y) = {y}^{2} - y + 1$

$p(0) = {0}^{2} - 0 + 1 = 1$

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

$p(t) = 2 + t + {2t}^{2} - {t}^{3}$

$p(0) = 2 + 0 + 2 \times {0}^{2} - {0}^{3} = 2$

$p(1) = 2 + 1 + 2 \times {1}^{2} - {1}^{3} = 4$

$p(2) = 2 + 2 + 2 \times {2}^{2} - {2}^{3} \\ \\ = 4 + 8 - 8 \\ \\ = 4$

$p(x) = {x}^{3}$

$p(0) = 0$

$p(1) = 1$

$p(2) = 8$

$p(x) = (x - 1)(x + 1)$

$p(0) = ( - 1)(1) \\ \\ = - 1$

$p(1) = (1 - 1)(1 + 1) \\ \\ = 0$

$p(2) = (2 - 1)(2 + 1) \\ \\ = 3$