Question

if x-a is a factor of x^6-ax^5+x^4-ax^3+3x-a+2, then find the other roots

Answer 1

x-a is a factor of $x^{6}-ax^{5}+x^{4}-ax^{3}+3x-a+2$ .
Here, the degree of polynomial is = 6.So, this polynomial has 6 roots.
P(a) = 0 ( If we insert a in place of x we get 0 remainder , because a factor                          completely divides it divident )
x-a = 0
x = a
Now,
Insert x at the place of a.
so,
$x^{6}-ax^{5}+x^{4}-ax^{3}+3x-x+2$ = 0
$x^{6}-x^{6}+x^{4}-x^{4}+3x-x+2$ = 0
2a - 2 = 0
2a = 2
a = $\frac{2}{2}$
a = 1
Therefore , first root is x - 1.

Now, to find second root divide x-1 by $x^{6}-ax^{5}+x^{4}-ax^{3}+3x-a+2$
You will have a quotient.
Now, again insert x = a and find the value of a .
Whatever will be the value would be the value of a.

I hope it helps .

Answer 2

Answer:

                       -2

Step-by-step explanation:

By the definition of a factor, the statement in the question means

x6−ax5+x4−ax3+3x−a+2=(x−a)P(x)  

where  P(x)  is a polynomial of  x  

Setting  x=a  in this equation,

a6−a6+a4−a4+3a−a+2=0×  P(x)  

a+2=0  

a=   −2