Question

answer the question fast

Answer 1

Solution:

Given: to test for divisiblity between $p(x)=(x-1)^{2a}-x^{2a}+2x-1$ with $q(x)=2x^3-3x^2+x$.

If the factors of q(x) are the factors of p(x),then, we can say it is divisible.
(12 is divisible by 4 as,
12=2x2x3
4=2x2  ... has same factors so,it is divisible.
same idea is can be applied to solve this problem.)

By applying remainder theorem,

$q(x)=2x^3-3x^2+x$
if x-0=0
x=0 is a factor,then
$q(0)=2(0)^3-3(0)^2+0$
$q(0)=2(0)-3(0)+0$
$q(0)=0$ ... ∴ (x) is a factor.

when, x-1=0
x=1

$q(1)=2(1)^3-3(1)^2+1$
$q(1)=2-3+1$
$q(1)=0$ ... ∴(x-1) is a factor

when, 2x-1=0
2x=1
$x= \frac{1}{2}$

$q( \frac{1}{2} )=2( \frac{1}{2} )^3-3 (\frac{1}{2} )^2+( \frac{1}{2} )$
$q( \frac{1}{2} )=2( \frac{1}{8} )-3( \frac{1}{4} )+( \frac{1}{2})$
$q( \frac{1}{2} )=( \frac{1}{4})-3( \frac{1}{4} ) +( \frac{2}{4} )$
$q( \frac{1}{2} )= \frac{1}{4} + \frac{2}{4} - \frac{3}{4}$
$q( \frac{1}{2} )=0$...∴ (2x-1) is a factor

∴$2x^3-3x^2+x=(x-1)(2x-1)(x)$

$p(x)=(x-1)^{2a}-x^{2a}+2x-1$

when x-0=0
the, x=0

$p(0)=(0-1)^{2a}-(0)^{2a}+2(0)-1$
[tex]p(0)=((-1)^2)^a-0+0-1 [/tex]
$p(0)=(1)^a-1$
$p(0)=1-1=0$....∴(x) is a factor

when x-1=0
x=1

$p(1)=(1-1)^{2a} -(1)^{2a}+2(1)-1$
$p(1)=(0)^{2a}-1+1$
$p(1)=0+0=0$...∴ (x-1) is a factor

when 2x-1=0
2x=1
x=1/2

$p( \frac{1}{2} )=(( \frac{1}{2} )-1)^{2a}-( \frac{1}{2} )^{2a}+2( \frac{1}{2} )-1$
[tex]p( \frac{1}{2} )=(- \frac{1}{2} )^{2a}-( \frac{1}{2} )^{2a}+1-1 [/tex]
$p (\frac{1}{2} )=( \frac{1}{4}^a )-( \frac{1}{4}^a )+0=0$...(2x-1) is a factor,
hece, all thefactors of q(x) are in p(x),it is divisible.

                                       Hope it Helps!!