Find the value of k if (a+b) is a factor of a^2(b+c)+b^2(c+a)+c^2(a+b) + k·abc and with this value of k factorize the expression.

Answer 1

You use "long division" of polynomials and make the remainder = 0.

First, you write the polynomial as if there is only one variable, say "a":

(b+c)a² + (b²+c²+kbc)a +(b²c+bc²)

and divide it by (a+b)

After a very tedious long division problem (it is where mathematicians like to say - "it is easy to see"),

the remainder will be (2b²c - kb²c)

It has to be equal 0 - there should not be a "remainder", since a+b was a factor.

So there you have it: k=2

The factorization will follow from the "long division" dividend - which will be very easy indeed.

Please contact me if you need details on the long division part.

Answer 2

It is (A+b)(a+C)(B+c) with k=2

Answer 3

Distributive property on the given expression:

a^2b + a^2c + b^2c + ab^2 + ac^2 + bc^2 + kabc

Compares that with:

(a+b)(a+c)(b+c) =

(a^2 + ac + ab + bc)(b + c) =

a^2b + abc + ab^2 + b^2c + a^2c + ac^2 + abc + bc^2 =

a^2b + ab^2 + b^2c + a^2c + ac^2 + bc^2 + 2abc

They are exactly the same, term by term, which forces k=2

Therefore, the factorization is (a+b)(a+c)(b+c)