# help solve 3

#### andy69

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.

#### undefined

MHF Hall of Honor
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.

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).

#### andy69

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).
so after its difference between the squares x^2-a^2 do you expand it with (x-a)

#### undefined

MHF Hall of Honor
so after its difference between the squares x^2-a^2 do you expand it with (x-a)
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

$$\displaystyle r=-1$$

$$\displaystyle s=-\alpha^2$$

$$\displaystyle t=\alpha^2$$

#### andy69

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

$$\displaystyle r=-1$$

$$\displaystyle s=-\alpha^2$$

$$\displaystyle t=\alpha^2$$
(i) r=1
(i)s +t=0

#### undefined

MHF Hall of Honor