Question

Find the polynomial P(x) with real coefficients having the specified degree, leading coefficients, and zeros.
Degree: 5
Leading Coefficient: 7
Zeros: 5 (multiplicity 2), 1, 3-i

Answer 1

you need 5 zeros.

Answer 2

that's all the problem states

Answer 3

oh, by the imaginary root theorem, 3+i is also a root of the equation
P(x)=7(x-5)(x-5)(x-1)(x^2-6x+10)=7x^5-119x^4+777x^3-2415x^2+3500x-1750

Answer 4

the last (x^2-6x+10) is because that is the quadratic equation that has a solution of 3-i and 3+i

Answer 5

Thanks, I would've never found the answer by myself

Answer 6

so i did (x-3+i)(x-3-i)

Answer 7

just rummaging through my algebra 2 book :D

Answer 8

This is a tough one alright. The furthest I can get is a polynomial of degree 4, with (some) complex coefficients:
\[x^4+(-14+i)x^3+(68-11i)x^2+(-130+35i)x+75-25i\]You also know that P(x) begins with \(7x^5\), so your final factor will be \((7x+?)\), which you can figure out by dividing \(7x^5\) by the degree-4 polynomial. I suspect that the root will be complex, given that P(x) has real coefficients, so the factor will be something like \((7x+j+ki)\). I think that if you multiplied that in to the degree-4 polynomial, you would end up with some massive polynomial involving j and k that you could use to solve for j and k. For instance, the first part would be like \[(7x+j+k)(x^4-14x^3+ix^3+68x^2-11ix^2 ...)\]\[=7x^5-98x^4+7ix^4+jx^4-14jx^3+jix^3+kix^4-14kx^3\]\[+kix^3+476x^3-77ix^3 ...\]By grouping all the \(x^4\) terms together, we get\[(-98+7i+j+ki)x^4\]\[-98+7i+j+ki \in \mathbb{R}\]I think that means k=-7. Not sure about that, but worth trying. Grouping all the \(x^3\) terms together:\[(-14j+ji-14k+ki+476-77i)x^3\]\[14j+ji-14k+ki+476-77i \in \mathbb{R}\]We need the i terms to equal zero in the end, so\[j+k-77=0\]\[j+(-7)-77=0\]\[j=84\]so the last factor is \((7x+84-7i)\), meaning the last zero is found at \(x=-12+i\). Now you can work your way up to the final polynomial, yay! (Tip: start with the degree 4 one, rather than going through it all again).

Answer 9

Woow!... Dalvoron(Y)

Answer 10

I should probably have arrived at the same answer as anhhuyalex, so I suspect there's some calculation error in there somewhere.

Answer 11

Yarp, I forgot to add the i in \(7x+j+ki\), which changes the \(x^3\) terms to \[(-14j+ji-14ki-k+476-77i)x^3\], which means that the i terms are\[j-14k-77=0\]\[j-14(-7)-77=0\]\[j=-21\] so the factor is \((7x+21-7i)\), and the root is \((x=3+i)\)