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Mathematics 11 Online
OpenStudy (anonymous):

The polynomial function P(x) = x^3 + bx^2 + cx + 2 has two distinct positive rational zeroes. Find b and c.

Parth (parthkohli):

\[x^3 + bx^2 + cx + 2 = (x - \alpha)(x - \alpha)(x - \beta)\]This would be the general approach. One root must repeat twice, if you have two roots.

OpenStudy (anonymous):

But I thought they were supposed to be distinct? @ParthKohli

Parth (parthkohli):

Alpha and beta are distinct. Because you have a polynomial with degree three, and the roots are rational, one root must repeat.

OpenStudy (anonymous):

Oh okay. What next? @ParthKohli

Parth (parthkohli):

I hope you get my point. x^2 - 10x + 25 is a polynomial of degree 2, but it has only one root that repeats twice.

Parth (parthkohli):

Now, expand the right-hand side.

Parth (parthkohli):

That is, \((x - \alpha)(x- \alpha)(x - \beta)\).

OpenStudy (anonymous):

\[x ^{3} - (\beta+2\alpha)x ^{2} + (2\alpha \beta + \alpha^{2})x - \alpha^{2}\beta \]

Parth (parthkohli):

Seems correct! Now compare coefficients.

OpenStudy (anonymous):

How do I do that?

Parth (parthkohli):

According to your polynomial, the constant term is 2. But we got \(\alpha^2 \beta\) in our expansion. So these two are equal.

OpenStudy (anonymous):

Okay, what next? @ParthKohli

Parth (parthkohli):

It is the same with coefficients of \(x\) and. \(x^2\).

Parth (parthkohli):

After solving, you get. \(\alpha = 1\) and. \(\beta = 2\).

Parth (parthkohli):

Can you do it now?

OpenStudy (anonymous):

But the answer here is b = -2 and c = -1....

Parth (parthkohli):

That is not the final answer.

Parth (parthkohli):

Compare the coefficients of \(x\). They must be equal.

OpenStudy (anonymous):

I don't understand. @ParthKohli

Parth (parthkohli):

I'm really sorry. You know that \(x^3 + bx ^2 + cx + 2 = x^3 -(\beta + 2\alpha) x^2 + (2\alpha \beta + \beta ^2) x + \alpha^2 \beta\)

Parth (parthkohli):

Let me come on my laptop.

OpenStudy (anonymous):

@ParthKohli Still need help :(

Parth (parthkohli):

Oh, hello!

Parth (parthkohli):

Where are you stuck?

OpenStudy (anonymous):

There when yu said that alpha = 1 and beta = 2...

Parth (parthkohli):

OK. Are you convinced that \(c = 2\alpha \beta + \beta^2\)?

OpenStudy (anonymous):

Yes.

Parth (parthkohli):

\[\alpha = 1, \beta = 2\]

Parth (parthkohli):

Use that to find \(c\).

Parth (parthkohli):

Hmm -- weird... the answer is not coming out as intended.

Parth (parthkohli):

Do you know the Rational Root Theorem?

OpenStudy (anonymous):

I got it! Thank you so much! @ParthKohli

Parth (parthkohli):

Oh... hmm! You're welcome. What is the final answer?

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