Question

factorise the polynomial- a(b-c) —b(c-a)​

Answer 1

Answer:

We first write the left hand side as a single fraction,

1(2−3⎯⎯√)3+1(2+3⎯⎯√)3=(2+3⎯⎯√)3+(2−3⎯⎯√)3(2−3⎯⎯√)3(2+3⎯⎯√)3,

What is the denominator? We have

(2−3⎯⎯√)3(2+3⎯⎯√)3=((2−3⎯⎯√)(2+3⎯⎯√))3=(4−3)3=13=1.

So our fraction becomes

(2+3⎯⎯√)3+(2−3⎯⎯√)3.

We can use the binomial theorem to write

(a+b)3(a−b)3=a3+3a2b+3ab2+b3=a3−3a2b+3ab2−b3,

so that we have (a+b)3+(a−b)3=2a3+6ab2. Thus

(2+3⎯⎯√)3+(2−3⎯⎯√)3=2×23+6×2×3=52,

as required.

Show that (a−b) is a factor of

a3(b−c)+b3(c−a)+c3(a−b)

Let’s define Q(a,b,c)=a3(b−c)+b3(c−a)+c3(a−b).

We can show that (a−b) is a factor of this expression using the Factor Theorem.

This says that (x−p) divides the polynomial P(x) if and only if P(p)=0.

So if we regard Q as a polynomial in a, then (a−b) will divide Q(a,b,c) if and only if Q(b,b,c)=0.

Now Q(b,b,c)=b3(b−c)+b3(c−b)+c3(b−b)=0, so (a−b) is a factor of Q(a,b,c).

Answer 2

Answer:

2ab+c(a-b)

Step-by-step explanation:

= ab-ac-bc+ab

= 2ab+c(a-b)