Question

Which of the following is a polynomial with roots – square root of 3, square root of 3, and –2 ?

x3 – 2x2 – 3x + 6

x3 – 3x2 – 5x + 15

x3 + 2x2 – 3x – 6

x3 + 3x2 – 5x – 15

Answer 1

if you mulitiply (x+a)(x+b) you get x^2 + (a+b)x +ab
if you multiply that by (x+c), the last term will be abc

if a root is given, e.g. x= \( \sqrt{3} \) then the corresponding factor is
\[ x= \sqrt{3} \\ (x-\sqrt{3})= 0\]
in other words, your polynomial is an expanded form of
\[ (x-\sqrt{3})( (x- -\sqrt{3}) (x--2) = 0 \\
(x-\sqrt{3})( (x+\sqrt{3}) (x+2) = 0
\]
now multiply out
\[ -\sqrt{3}\cdot \sqrt{3}\cdot 2 \]
to get the last term

Answer 2

@phi is there something i'm supposed to do with imaginary numbers?

Answer 3

are the roots imaginary ? in other words are they
\[ \sqrt{3}\ i \text{ or }\sqrt{-3}\]?

Answer 4

@phi I thought that we could replace the root with the imaginary i. Can we not do that?

Answer 5

A "root" of an equation is a number that makes the equation zero.
They gave you the 3 roots. They are
\[ x= \sqrt{3},\ x= - \sqrt{3} , \ x= -2 \]

Answer 6

as posted up above, you can re-write each of those as
\[ (x - \sqrt{3}) = 0 \\(x + \sqrt{3}) = 0 \\(x+2) = 0 \]
if you multiply them together you get
\[ (x - \sqrt{3}) (x + \sqrt{3})(x+2) = 0 \]
that is the "factored" form of your equation. It should be obvious that if x is any of the "root" values, you will get a 0 in one of the 3 terms, and 0 times the rest gives 0.

If you multiply out the 3 terms you will get one of the 4 choices. I was trying to show you a short cut to multiplying it out... the last term in the expanded equation will be the product of the roots.

or you can multiply it out and see what you get.

Answer 7

sorry, I guess I've been staring at numbers for too long now.

So, (x - √9) (x+2) ?

Answer 8

I mean (x^2 - √9) (x+2)

Answer 9

and sqrt(9) is 3

Answer 10

Oh! Okay
So (x^2 - 3) (x+2)
(x^3 + 2x^2 -3x -6)

Answer 11

yes. looks good.

the short cut was to notice that the -6 comes from
\[ -\sqrt{3} \cdot \sqrt{3} \cdot 2 = -6 \]
and only one of the choices has -6... but multiplying it out is OK if you have the time.

Answer 12

btw, notice that to "solve"

x^3 + 2x^2 -3x -6 = 0
you would "go backwards" and try to factor it into
\[ (x - \sqrt{3}) (x + \sqrt{3})(x+2) = 0 \]
that is a much harder operation. It is relatively easy to go from factored to expanded...

Answer 13

Thanks for putting up with me. (: I appreciate it.