The cubic polynomial P(x) =x3 +rx2 +sx +t
, where r, s and t are real numbers, has three real zeros, 1, αand –α.
(i) Find the value of r.
(ii) Find the value of s +t.
P(x) =x3 +rx2 +sx +t.
with working out please.
This is not as hard as it may look.
Write $\displaystyle P(x) = (x-1)(x-\alpha)(x-(-\alpha))$, expand, and then match the coefficients.
Note that $\displaystyle (x-\alpha)(x+\alpha)=x^2-\alpha^2$ (in other words, you can skip the step of writing out that expansion explicitly by noticing there is a difference of squares).
I mentioned the difference of squares as a "shortcut."
Normally one might write
$\displaystyle (x-\alpha)(x+\alpha) = x^2 + x\alpha - x\alpha - \alpha^2 = x^2 -\alpha^2$.
Recognising a difference of squares allows you to skip the middle step and instantly jump to the last step.
Anyway, what you'll want is
$\displaystyle P(x)=(x-1)(x^2-\alpha^2)= x^3 - x^2 -\alpha^2x + \alpha^2$
which allows you to write
$\displaystyle r=-1$
$\displaystyle s=-\alpha^2$
$\displaystyle t=\alpha^2$