Question

Polynomial with roots 4, -2i, and 2i?

Answer 1

If you have the roots \(r_1,r_2,...r_n\) , you can construct a polynomial with those roots:
\[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is a constant 1, or any value needed to make the polynomial pass through an arbitrary point.

Answer 2

So I must plug in the numbers 4, -2, and 2 into p(x) = a(x - r1)(x - r2)(x - rn)?

Answer 3

Things to remember about this: if you have a polynomial with only real coefficients, any complex roots (such as \(-2i\)) will come in complex conjugate pairs \(a \pm bi\).

You're probably expected to multiply the whole thing out. It will be easier if you multiply the parts with the complex roots first.

Answer 4

No, use \(4, -2i, 2i\)...

Answer 5

I must warn you that Math is not my forte. :P

Answer 6

ok so it would be: p(x) = a(4 - r1)(-2i - r2)(2i- rn)?

Answer 7

No. You substitute the roots for the \(r_1,r_2,...\), not the \(x\)

For example, polynomial with roots \(1,-2\) would be
\[P(x) = a(x-1)(x- (-2)) = a(x-1)(x+2) = a(x^2+x-2) \]\[\qquad= ax^2 + ax -2a\]
And probably we would take \(a = 1\) so that becomes
\[P(x) = x^2+x-2\]

Quick plot shows we cross the x-axis at\( x = -2,\, x = 1\), just like we expect if our roots are \(1, -2\)

Answer 8

thank you so much by the way. :)

Answer 9

Notice that the value of \(a\) plays absolutely NO role in the location of the roots. Here are a bunch of different polynomials based on varying \(a\) in my example:

Answer 10

Alright so the answer would be: p(x) = x^3 - 4x^2 + 4x - 16

Answer 11

Magic 8-ball says "Yes"

Answer 12

Yay! Thank you whpalmer4!

Answer 13

I would say "an answer"...as I tried to show, you can have as many polynomials as you want with those same roots. \(P(x) = 2x^2-8x^2+8x-32\) is equally valid (multiplied each term by 2)

Answer 14

Sometimes these problems will be "find a poly with roots blah blah that goes through the point (x,y)"

in which case you find \[P(x) = ax^2 +...\]and you plug in the value of the point and solve for \(a\) so that P(x) = the specified value of y...

Answer 15

A somewhat less common variant: polynomial with roots blah blah such that the leading coefficient is some value. Again, same procedure.

Answer 16

One of the reasons I like math: learn one concept, unlock a tool for doing many different problems. Compare with history: learn one fact, know one fact :-)

Answer 17

Well you should be incredibly proud of yourself because you are obviously exceptionally smart. :) Thank you for going the extra mile and giving me that extra information! :) Have a good day!

Answer 18

You're welcome!