It is given that x3 +bx2 +cx+d ≡ (x−α)(x−β)(x−γ). By substituting 3 different values of x show that:
αβγ = −d
(1 − α)(1 − β)(1 − γ) = 1 + b + c + d
(1 + α)(1 + β)(1 + γ) = 1 − b + c − d
where α, β and γ are integers, show using one of the equations from part (iv) that:
x3 −4x2 −4x+16≡(x−α)(x−β)(x−γ), (1 + α)(1 + β)(1 + γ) = −15 .
What are the 8 possible values of α? Find the possible values of 1 + α first. Write down the value of αβγ. Which of the 8 values of α can you now eliminate?
Now use the remaining equation from part (iv) to bring the possible values of α down to three.
Someone pls help with this question I’m stuck. The hints and partial solutions says that to work out 1 + alpha, you need to work out the factors of 15 but I don’t understand why. Also, I don’t get which of the 8 values of alpha to eliminate. Pls help