Math, asked by dhruvanshpanjwani, 1 year ago

if 0 and 1 are the zeroes of the polynomial F(X)=2x^3-3x^2+ax+b. find the value of a and b​

Answers

Answered by abhi569
59

It is given that both 0 and 1 are the zeroes of the polynomial f(x) = 2x^3 - 3x^2 + ax + b, so numeric value of the polynomial should be 0 on substituting the numeric value of zeroes.

Therefore, f(0) = 0

= > 2(0)^3 - 3(0)^2 + a(0) + b = 0

= > 2(0) - 3(0) + a(0) + b = 0

= > b = 0

Hence the required value of b is 0

When x = 1, f(1) = 0

= > 2(1)^3 - 3(1)^2 + a(1) + b = 0

= > 2(1) - 3(1) + a( 1 ) + 0 = 0 {b=0}

= > 2 - 3 + a = 0

= > - 1 + a = 0

= > a = 1

Hence, a = 1 & b = 0


Anonymous: Bhai !! 0ⁿ = 1
siddhartharao77: 0^n = 0 {n > 0}
abhi569: 0ⁿ , substitute n = 3 , 0³ = 0 x 0 x 0 = 0 , substitute n = 2 , 0² = 0 x 0 = 0 , etc......thus 0ⁿ = 0
Answered by Anonymous
28

Polynomial :

f(x) = 2x³-3x²+ax+b

Given zeroes of the f(x) = 0 and 1 .

Put x = 0 in f(x) :

= 2x³-3x²+ax+b

= 2(0)³-30²+a0+b

= 2×0-3×0+0+b

= 0+b

Consider this = 0

b = 0

Then ,

put x = 1 in f(x)

= 2x³-3x²+ax+b

= 2(1)³-3(1)²+a×1+b

= 2-3+a+b

= -1+a+b

Consider this = 0

-1+a+b = 0

Put b = 0 (Obtained above)

-1+a+0= 0

a = 1

The required values of a and b are 1 and 0 respectively.


siddhartharao77: Please, recheck it dear!
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