Q13/120) The polynomial p(x)=x^3-3ax^2+bx-6 has factors (x-1) and (x-3). Find the value of the constants a and b?

This can be solved using  factor theorem and putting p(1) = 0 and p(3) = 0.

As below

p(1) = 1 – 3a + b – 6 = 0 or -3a + b = 5  … (1)

p(3) = 27- 27 a + 3b – 6 = 0 or 27 a -  3b = 21 … (2)

we can solve above 2 linear equations to get a = 2 and b = 11

But there is a shorter method

As (x-1) and (x-3) are factors so

P(x) = m(x-1)(x-3)(x-n) = x^3-3ax^2+bx-6

Now coefficient of x^3 = m = 1

Constant term = -3mn = -6 or n = 3

So f(x) = (x-1)(x-2)(x-3) = x^3 – 6x^2 + 11x – 6

comparing coefficients we get a = 2 and b = 11