Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!

Algebra ->  Algebra  -> College  -> Linear Algebra -> Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!     (Log On)

   

Question 166514: For x²(a-b)+a²(b-x)+b²(x-a) show that (x-a) is a linear factor.
PLEASE HELP ME!!!!!!: For x²(a-b)+a²(b-x)+b²(x-a) show that (x-a) is a linear factor.
PLEASE HELP ME!!!!!!
Answer by Edwin McCravy(2199) About Me  (Show Source):
You can put this solution on YOUR website!
For x^2(a-b)+a^2(b-x)+b^2(x-a) show that (x-a) is a linear factor.

We use the fact that if a number is substituted for x 
in a polynomial, the result is the same as the remainder 
left when that polynomial is divided by x minus that number.

So we substitute x=a in the polynomial to see if the
remainder is 0, and if it is then we'll know that if we 
had divided the polynomial by x-a, the reaminder
would be 0, making x-a a factor:

So we substitute x=a into the polynomial:

x^2(a-b)+a^2(b-x)+b^2(x-a)
a^2(a-b)+a^2(b-a)+b^2(a-a)
a^3-a^2b+a^2b-a^3+b^2(0)
a^3-a^2b+a^2b-a^3+0

and everything cancels out and we have

0

so 0 would be the remainder if 

x^2(a-b)+a^2(b-x)+b^2(x-a)

were divided by x-a, so 

x-a is a factor of  x^2(a-b)+a^2(b-x)+b^2(x-a).

And of course x-a is linear because the highest power
of x is 1.

Edwin
Question 166514: For x²(a-b)+a²(b-x)+b²(x-a) show that (x-a) is a linear factor.
PLEASE HELP ME!!!!!!: For x²(a-b)+a²(b-x)+b²(x-a) show that (x-a) is a linear factor.
PLEASE HELP ME!!!!!!
Answer by ptaylor(1368) About Me  (Show Source):
You can put this solution on YOUR website!
First, get rid of parens
x^2a-x^2b+a^2b-a^2x+b^2x-b^2a rearrange and regroup
(a-b)x^2-(a^2-b^2)x+(a^2b-b^2a) expand (a^2-b^2)
(a-b)x^2-(a-b)(a+b)x+ab(a-b) factor out (a-b)
(a-b)(x^2-(a-b)x+ab)=
Since x^2-(a-b)x+ab equals (x-a)(x-b), we have
(a-b)(x-a)(x-b)

Hope this helps---ptaylor