1. ## help solve 3

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.

2. Originally Posted by 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.

This is not as hard as it may look.

Write $P(x) = (x-1)(x-\alpha)(x-(-\alpha))$, expand, and then match the coefficients.

Note that $(x-\alpha)(x+\alpha)=x^2-\alpha^2$ (in other words, you can skip the step of writing out that expansion explicitly by noticing there is a difference of squares).

3. Originally Posted by undefined
This is not as hard as it may look.

Write $P(x) = (x-1)(x-\alpha)(x-(-\alpha))$, expand, and then match the coefficients.

Note that $(x-\alpha)(x+\alpha)=x^2-\alpha^2$ (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)

4. Originally Posted by andy69
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

$(x-\alpha)(x+\alpha) = x^2 + x\alpha - x\alpha - \alpha^2 = x^2 -\alpha^2$.

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

$P(x)=(x-1)(x^2-\alpha^2)= x^3 - x^2 -\alpha^2x + \alpha^2$

which allows you to write

$r=-1$

$s=-\alpha^2$

$t=\alpha^2$

5. Originally Posted by undefined
I mentioned the difference of squares as a "shortcut."

Normally one might write

$(x-\alpha)(x+\alpha) = x^2 + x\alpha - x\alpha - \alpha^2 = x^2 -\alpha^2$.

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

$P(x)=(x-1)(x^2-\alpha^2)= x^3 - x^2 -\alpha^2x + \alpha^2$

which allows you to write

$r=-1$

$s=-\alpha^2$

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

6. Originally Posted by andy69
(i) r=1
(i)s +t=0
Almost. r = -1, not 1. I hope you see how I matched the coefficients, because that's the main point of the problem, and you might get more problems like it.

7. Originally Posted by undefined
Almost. r = -1, not 1. I hope you see how I matched the coefficients, because that's the main point of the problem, and you might get more problems like it.
yeah thanks i forgot the minus,great help anyway