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Polynomial ax^3+3x^2-3 and 2x^3-5x+a when divided by x-4 leave the same remainder. find the value of a.
p(x1)=ax3+3x2-3 & p(x2)=2x3-5x+a

As (x-4) is the zero of the polynomial so, x=4

Now putting the value of 'x ' in the polynomials,

p(4)=a*43+3*42-3 & p(4)=2*43-5*4+a

As both the Eqn. have the same remainder so,

p(x1)=p(x2)

a*43+3*42-3 = 2*43-5*4+a

64a+48-3 = 128-20+a

64a-a=108-45

63a=63

a = 1


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