if the polynomials 2xcube+ ax square+3x-5 and xcube+x square-2x+a leave the same remainders when divided by(x-2) find the value of a
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x-2 = 0
X = 2
P(2) = 2*(2)^3 + a*(2)^2 + 3*2 - 5
= 16 + 4a + 6 - 5 = 17+4a (1)
P(2) = (2)^3 + (2)^2 - 2*2 + a
= 8 + 4 - 4 + a = 8+a (2)
Their remainders are equal
17+4a = 8+a
3a = - 9
a = - 3
X = 2
P(2) = 2*(2)^3 + a*(2)^2 + 3*2 - 5
= 16 + 4a + 6 - 5 = 17+4a (1)
P(2) = (2)^3 + (2)^2 - 2*2 + a
= 8 + 4 - 4 + a = 8+a (2)
Their remainders are equal
17+4a = 8+a
3a = - 9
a = - 3
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