Question

the polynomials px3+2x-3 and x2-px+4 leaves the same remainder.Find the value of p​

Answer 1

Answer

The Remainder is same whne (x−3) divides (x3−px2+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

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Answer 2

Answer

The Remainder is same whne (x−3) divides (x3−px2+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

HOPE IT HELPED you

same as up our both answer correct 100%

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