Question

Factorise : a(a+1)x^2-x-a(a-1)​

Answer 1

Answer:

a2x2 - a2 - ax2 - a + x

Step-by-step explanation:

Equation at the end of step  1  :

 (((a•(a-1))•(x2))+x)-a•(a+1)

Step  2  :

Equation at the end of step  2  :

 ((a•(a-1)•x2)+x)-a•(a+1)

Step  3  :

Equation at the end of step  3  :

 (ax2 • (a - 1) +  x) -  a • (a + 1)

Answer 2

Thanks for the chance to answer this question.

We know that for any given quadratic equation ax^2 + bx + c = 0, the roots of the equation can be calculated by the Sridhar Acharya formula as given below -

x = [-b +/- (b^2 - 4ac)^1/2] / 2a -------- (1)

Now, let us assume that the given polynomial is equal to zero for certain values of x. If we are able to find out those values, then the whole polynomial can be represented as a combination of the solution polynomials.

Hence, a(a+1)x^2 - x - a(a-1) = 0, for some values of x.

Applying the formula (1) to the above equation, we see that -

A = a(a+1)

B = -1

C = -a(a-1)

Hence, x = [-(-1) +/- {(-1)^2 - 4a(a+1)*(-a(a-1))}^1/2] / 2a(a+1)

=> x = [1 +/- {1 - 4a^2(a^2 - 1)}] / 2a(a+1)

=> x = [1 +/- (2a^2 - 1)] / 2a(a+1)

From the above equation, we get 2 values of x as mentioned below -

x = a / (a+1) and x = (1-a) / a

=> ax - a + x = 0 and ax + a - 1 = 0.

So, for these two values of x, the equation is justified, or the given polynomial is a combination of the above two simpler polynomials.

That means, a(a+1)x^2 - x - a(a-1) = (ax - a + x)(ax + a - 1)