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.

Thank u so much. U really made my day. With deep appreciation

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?

In case, I don’t understand Amaths qns, May ask u guys? My name is Vignesh My h/p 88173761. If u wish to help pls pm me. Thanks

Thank u very much. I deeply appreciate it. Thanks again

Vignesh, for the guessing you can just use your in-built calculator function. Also, you must be prepared that the questions can require topics from more than one chapter.

For some reason I can’t see the working fully (it got cut off halfway when I view it from my phone).

a³ = 4 - 3a

a³ + 3a - 4 = 0

let f(a) = a³ + 3a - 4

Sub a = 1,

f(1) = 1³ + 3(1) - 4
= 1 + 3 - 4
= 0

By the factor theorem, (a - 1) is a factor of f(a)


So (a - 1)(a² + a + 4) =0

You can easily guess the coefficients of a for the quadratic factor since we know only a² x a = a³ and only 4 x -1 = -4.

Otherwise, use long division


a = 1 or a² + a + 4 = 0

For a² + a + 4 = 0,

Discriminant (b² - 4ac)

= 1² - 4(1)(4)
= 1 - 16
= -15 < 0

Since discriminant < 0, there are no real roots for this part

∴ a = 1

You have to show there are no real roots

Thank u very much

@ J, I find your method much easier to understand for finding the value of a

@ Eric, u r right . I used the calculator and put a^3+3A-4. I got a=1 and other 2 roots r not real. Thanks

@ Raymond, thanks a lot for all the help

Thanks guys

My calculator model is Casio FX 96 SG Plus. There is an in built function there MODE —> 3 —> 4: aX3 + bX2 + cX + d or something like that. Any non-real roots will have a letter “lowercase i” at the end. So if you ever see two “i” (they come in twos for cubic functions), these are not real.

Yes. I used to have that model. Now using Casio fx-97SG X. This one can find roots until x to the power of 4.

Do note that while you may use the calculator to find the roots, the working still has to be shown as above.

Sure. Will do that. Thanks a lot for all the guidance