Question

Find the value of a for which the polynomials x3+x+6 and 2x3-x2-(a+3)x-6, leave the same remainder when divided by x-3

Answer 1

Answer:

The Remainder is same whne (x−3) divides (x

3

−px

2

+x+6) & (2x

3

−x

2

−(p+3)x−6)

∴ Using Remainder Theorem

R(3)=x

3

−px

2

+x+6

=3

3

−p(3

2

)+3+6

=27−9p+3

=36−9p

R(3)=2x

3

−x

2

−(p+3)x−6

=2(3

3

)−3

2

−(p+3)3−6

=2×27−9−3p−9−6

=54−24−3p

=30−3p

Remainder are same

∴36−9p=30−3p

36−30=−3p+9p

6=6p

1=p

Stay home and stay safe :-)

Answer 2

let, p( x1) = ( x-3) × q(x1) + r( x)

so, from formula p(a) = r(x)

we get p(3) = r(x)

r(x) = 27+3+6 = 36

now,

p(x2) = (x-3) × q(x2) + r(x)

so, again from formula we get,

r(x) = p(x2)

36 = 54-9-3a-9-6

6 = ( -3a )

a = ( -2 )

Hope I am able to help you