help solve 3

May 2010
39
0
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.
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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).
 
May 2010
39
0
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)
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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

\(\displaystyle r=-1\)

\(\displaystyle s=-\alpha^2\)

\(\displaystyle t=\alpha^2\)
 
May 2010
39
0
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

\(\displaystyle r=-1\)

\(\displaystyle s=-\alpha^2\)

\(\displaystyle t=\alpha^2\)
so answering my questions
(i) r=1
(i)s +t=0
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
so answering my questions
(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.
 
May 2010
39
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.
yeah thanks i forgot the minus,great help anyway