Question

Which of the following is a polynomial with roots negative sq root 3, sq root 3, and 2?

Answer 1

If we have some polynomial factored out as

f(x) = (x-a)(x-b)(x-c)... = 0 then the roots are a,b,c...

For example, \[f(x) = x(x-1) = x^2 -x \]
The roots are x = 0 and x = 1, which you can easily verify:
\[0^2-0 = 0\]\[1^1-1= 0\]

So, make a factored polynomial from your roots, then multiply it out if the answer choices are not factored.

Answer 2

x3 + 3x2 - 5x - 15

@whpalmer4

Answer 3

Well, let's check that:
\[f(x) = x^3 + 3x^2 -5x - 15 = 0\]\[f(2) = 2^3 + 3(2^2)-5(2) - 15 = 8 + 3*4 - 10 = 10\]But f(2) = 0 if x=2 is a root!

Can you show me what you multiplied together to get your polynomial?

Answer 4

I incorporated the 3's because that's what was given in the problem

Answer 5

3x2 because that's the additional number used

Answer 6

5 x 3 = 15

Answer 7

Well, the devil is in the details. Can you show me what you multiplied, exactly?

Answer 8

would it be plus fifteen or minus? (positive/negative)?

Answer 9

Okay, we know the roots are
\[a=2, b = -\sqrt{3}, c = \sqrt{3}\]
Given the roots, the factored polynomial is

\[(x-a)(x-b)(x-c) = (x-2)(x-(=\sqrt{3})(x-\sqrt{3}) = (x-2)(x+\sqrt{3})(x-\sqrt{3})\] because the roots are the places where the (x-a,b,c,etc.) = 0. What you have to do is multiply that out to get something of the form x^3 + x^2 + x + c. You'll have different coefficients than all 1s and c, of course.

Answer 10

oh my god this is difficult

Answer 11

Well, it's tedious and error-prone, at least :-)

You might try just multiplying out (x-a)(x-b)(x-c) and then substitute a = 2, etc. into the resulting equation.

Answer 12

It MUST be +15

Answer 13

It's a negative radical in the beginning, sure but...
x3 - 3x2 - 5x + 15... positive towards the end

Answer 14

Or, notice that you've got \[(x-a)(x+a) = x^2 - ax + ax - a^2\] with the \[(x-\sqrt{3}),(x -(-\sqrt{3}))\]terms, which would give you (x^2-3) for that part.

\[(x-2)(x^2-3) = x^3-3x-2x^2+6=x^3-2x^2-3x+6\]
Let's check that equation out:
\[f(2) = 2^3-2(2^2)-3(2)+6 = 8 - 2(4) -3(2) +6 = 0\]So far, so good!
\[f(\sqrt{3}) = \sqrt{3}*\sqrt{3}*\sqrt{3} - 2(\sqrt{3}*\sqrt{3} - 3*(\sqrt{3}) + 6 = 3*\sqrt{3}-2(3)-3*\sqrt{3} + 6 = 0\]Another success!
You can verify the case for f(-sqrt(3)) yourself.

Answer 15

Doing it the harder way:
\[(x-2)(x-\sqrt(3))(x+\sqrt(3))=(x^2-\sqrt{3}x-2x-(-2)\sqrt{3})(x+\sqrt{3})=\]\[(x^2-\sqrt{3}x-2x+2\sqrt{3})(x+\sqrt{3}) = \]\[x^3+\sqrt{3}x^2-\sqrt{3}x^2-\sqrt{3}\sqrt{3}x-2x^2-2\sqrt{3}x+2\sqrt{3}x+2\sqrt{3}\sqrt{3}=\]\[x^3-3x-2x^2+2(3)=x^3-2x^2-3x+6\]Just like we got before.

Answer 16

thank you so much you're amazing!

Answer 17

I've had a lot of practice. With a lot of practice, you will also amazing :-)