polynomials ax^3-3x^2+4 and 2x^3-5x+a when divided by x-2 leave the remainders p and q respectively if p-2q=10 find the value of k
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x=2
:.ax^3-3x^2+4 = 8a-12+4= 8a-8
since,
ax^3-3x^2+4 = p
therefore,
8a-8=p
:. 2x^3-5x+a =16-10+a= a+6
since,
2x^3-5x+a =q
therefore,
a+6=q
now,
since,
p-2q=10
therefore,
8a-8-2a-12=10
6a-20=10
6a=30
a=5
:.ax^3-3x^2+4 = 8a-12+4= 8a-8
since,
ax^3-3x^2+4 = p
therefore,
8a-8=p
:. 2x^3-5x+a =16-10+a= a+6
since,
2x^3-5x+a =q
therefore,
a+6=q
now,
since,
p-2q=10
therefore,
8a-8-2a-12=10
6a-20=10
6a=30
a=5
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