Question

Find the remainder when xcube - px square +6x- p is divided by x-p​

Answer 1

Answer:

ANSWER

Here, p(x)=x

3

+ax

2

+6x−a, and the zero of x−a is a

So, p(a)=a

3

+a×a

2

+6×a−a

=2a

3

+5a

So, by the Remainder Theorem, 2a

3

+5a is the remainder when x

3

+ax

2

+6x−a is divided by x−a.

Answer 2

$\star{\mathbb{\underline{ANSWER:}}}$

$let \: p(x) = {x}^{3} - p{x}^{2} + 6x - p$

$Divisor \: = x – p$

$Zero \:of \: x – p = p$

${By \:remainder \:theorem,}$

$If \:p(x) \:is \:divided \:by \:(x – p), \:then \:the \:remainder \:is \:p(p)$

$p ( p) = {p}^{3} - p ({p})^{2}+ 6 (p) -p$

$={p}^{3}-{p}^{3}+6p- p$

$= 6p - p$

$=5p$