The polynomial function P(x) = x^3 + bx^2 + cx + 2 has two distinct positive rational zeroes. Find b and c.
\[x^3 + bx^2 + cx + 2 = (x - \alpha)(x - \alpha)(x - \beta)\]This would be the general approach. One root must repeat twice, if you have two roots.
But I thought they were supposed to be distinct? @ParthKohli
Alpha and beta are distinct. Because you have a polynomial with degree three, and the roots are rational, one root must repeat.
Oh okay. What next? @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.
Now, expand the right-hand side.
That is, \((x - \alpha)(x- \alpha)(x - \beta)\).
\[x ^{3} - (\beta+2\alpha)x ^{2} + (2\alpha \beta + \alpha^{2})x - \alpha^{2}\beta \]
Seems correct! Now compare coefficients.
How do I do that?
According to your polynomial, the constant term is 2. But we got \(\alpha^2 \beta\) in our expansion. So these two are equal.
Okay, what next? @ParthKohli
It is the same with coefficients of \(x\) and. \(x^2\).
After solving, you get. \(\alpha = 1\) and. \(\beta = 2\).
Can you do it now?
But the answer here is b = -2 and c = -1....
That is not the final answer.
Compare the coefficients of \(x\). They must be equal.
I don't understand. @ParthKohli
I'm really sorry. You know that \(x^3 + bx ^2 + cx + 2 = x^3 -(\beta + 2\alpha) x^2 + (2\alpha \beta + \beta ^2) x + \alpha^2 \beta\)
Let me come on my laptop.
@ParthKohli Still need help :(
Oh, hello!
Where are you stuck?
There when yu said that alpha = 1 and beta = 2...
OK. Are you convinced that \(c = 2\alpha \beta + \beta^2\)?
Yes.
\[\alpha = 1, \beta = 2\]
Use that to find \(c\).
Hmm -- weird... the answer is not coming out as intended.
Do you know the Rational Root Theorem?
I got it! Thank you so much! @ParthKohli
Oh... hmm! You're welcome. What is the final answer?
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