2
$\begingroup$

I know the factor theorem i.e,

Let $P(x)$ be a polynomial of degree greater than or equal to $1$ and $a$ be a real number such that $P(a) = 0$, then $(x-a)$ is a factor of $P(x)$.

I have an question in my textbook which is -

  • Using Factor theorem , show that $a-b,b-c$ and $c-a$ are the factors of $$a(b^2 -c^2)+b(c^2-a^2)+c(a^2-b^2).$$

I can not see any polynomial over here . How can I solve the problem ?

Hints are welcome

$\endgroup$
1
  • 1
    $\begingroup$ Replace $a$ with $X$. Find a factor. Replace $b$ with $X$. Find a factor. Replace ... $\endgroup$ Jul 10, 2015 at 10:52

2 Answers 2

2
$\begingroup$

Instead of a polynomial in $x$ or $y$ as usual, consider the expression $a(b^2 -c^2)+b(c^2-a^2)+c(a^2-b^2)$ to be firstly a polynomial in $a$, then in $b$ then $c$. Each use of the factor theorem on each case should give you one solution.

$\endgroup$
0
$\begingroup$

Hint: Using the difference of two squares we can factorise your expression as $$a(b^2 -c^2)+b(c^2-a^2)+c(a^2-b^2) = a(b-c)(b+c) + b(c-a)(c+a) + c(a-b)(a+b)$$

$\endgroup$
1
  • $\begingroup$ What comes next by this method ? $\endgroup$
    – 5cube
    Jul 10, 2015 at 11:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .