Step-by-step explanation: x^2 -1 =0 => x^2 = 1 => x = 1 also x^3 =1 (1×1×1 = 1) substituting this value in the polynomial we get: a×(1)^3 +(1)^2 -2×1 + b =0 = a + 1 – 2 +b =0 => a + -1 +b = 0 => a + b = 1 ( transposing 1 to RHS) Had the second equation being given, we would have got the values of a & b Reply
Answer: Step-by-step explanation: p(x)=ax^3+x^2+-2x+b quotient:- kuch kuch answer kuch kuch diy Reply
Step-by-step explanation:
x^2 -1 =0
=> x^2 = 1
=> x = 1
also x^3 =1 (1×1×1 = 1)
substituting this value in the polynomial we get:
a×(1)^3 +(1)^2 -2×1 + b =0
= a + 1 – 2 +b =0
=> a + -1 +b = 0
=> a + b = 1 ( transposing 1 to RHS)
Had the second equation being given, we would have got the values of a & b
Answer:
Step-by-step explanation:
p(x)=ax^3+x^2+-2x+b
quotient:- kuch kuch
answer kuch kuch
diy