Find p(0), p(1) and p(2) for each of the following polynomials; (i) p(y)= y²-y+1 (ii) p(t)=2+t+2t²-t^3 (iii) p(x)=x^3 (iv) p(x)=(x-1)(x+1) [ncert 9th maths]

      Excercise 2.2

Q2.

(i) Given that p(y)=y²-y+1

=> p(0)= 0²-0+1 =0+0+1=1 Ans.

p(1)=1²-1+1=1-1+1=0+1=1 Ans.

p(2)= 2²-2+1 =4-2+1=5-2=3 Ans.

(ii)  p(t)=2+t+2t²-t^3

p(0)=2+0+2(0)²-(0)^3

p(0)=2+0+0-0  

   [∵  0²=0×0 =0 and 0^3=0×0×0=0]

p(0)=2  Ans.

p(1)=2+1+2(1)²-(1)^3

p(1)=2+1-2(1×1)-(1×1×1)

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

p(1)=0 Ans.

P(2)=2+2+2(2)²-(2)^3

p(2)=2+2+2(2×2)-(2×2×2)

p(2)=2+2+8-8

p(2)=4 Ans.

(iii) Try yourself.

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

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

p(0)=(-1)(1)= -1 Ans.

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

p(1)=0×2

p(1)=0 Ans.

p(2)= - - - -      [Try yourself]

Find x , if 1/(9!) + 1/(10!) = x/(11!).

 

Here n! is read as n factorial. It can be written as n(n-1)! or n(n-1)(n-1)!

Expand the following using suitable identity (i) (x+2y+4z)² (ii) (-2x+3y+2z)². [ncert]

(i) (x + 2y + 4z)²  ( iii )  (- 2x + 3y + 2z) ^ 2

We know that  (a + b + c)² = a²+ b²+ c²+ 2ab + 2bc + 2ca

(i) (x + 2y + 4z)² = (x)² + (2y)²+ (4z)²+ 2(x)(2y)+2(2y)(4z) + 2(4z)(x)

= x² + 4y²+ 16z² + 4xy + 16yz + 8zx  Ans.

(ⅱ) (- 2x + 3y + 2z)²=(-2x)²+(3y)²+(2z)² +2(-2x)

(3y)+2(3y)(2z) + 2(2z)(- 2x)

 = 4x²+9y²+4z²-12xy+12yz-8zx  Ans.