In (a - 1) (a + 2) (a + 2),

(a + 2) is our repeated factor, while (a - 1) is our singular factor.

Somehow, when you learn imaginary and complex numbers in the A Levels (it’s the case for “no real roots”, which we can’t simply call it “no roots” because actually we have what we call “imaginary/complex roots”), you will realise that the repeated factor above will somehow lead to complex roots.

A cubic equation will always have three roots regardless of whether they are real or not, but in our case we only have two “real” solutions. There must be a third “non-real” root out there, and this comes from the repeated real factor.

The only “absolutely real” root which we can have is the non-repeated version, which is a = 1.

Hence, the value of our expression is 1.

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You have performed the long division incorrectly, it should be factored into (a-1)(a^2 + a + 4) which has 1 as a real root and two other complex roots. It can be easily checked that a = -2 does not satisfy the equation.

Thanks to both of u

Good point, I was mentally calculating it in my head and fully forgot that the number to achieve is -3x (I was thinking of it as 0x).

Even if it was correct, the repeated root (a + 2) is already real... How can it be non-real?

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