Question

Which of the following shows all the real roots of 4x^3 + 4x^2 -8x = 0

Answer 1

-2, 0, 1
-4, 0, 1
-2, 2, 4
-4, 0, 4

Answer 2

@ranga Help?

Answer 3

Factor out 4x first. Then quickly try each answer by substituting the numbers into the equation and see if it goes to 0.

Answer 4

how do i factor 4x?

Answer 5

4x^3 + 4x^2 -8x = 0
4x(x^2 + x - 2) = 0
4x(x+2)(x-1) = 0
Therefore, x = 0, 1 or -2

Answer 6

Do i plug them in now and try for each of those to get 0?

Answer 7

@ranga

Answer 8

x= 0 so how do i find the roots?

Answer 9

The roots are the values of \(x\) such that \(f(x) = 0\)

\[4(-2)^3+4(-2)^2-8(-2) = 4*(-8)+ 4*4 + 16 = -32+16+16 = 0\]Therefore \(x=-2\) is a root.
\[4(0)^3+4(0)^2-8(0) = 0+0-0=0\]Therefore \(x=0\) is a root.
\[4(1)^3+4(1)^2-8(1) = 4+4-8=0\]Therefore \(x=1\) is a root.

Answer 10

When you have\[ a*b*c = 0\]the only way that can be true is if one or more of \(a,b,c =0\), right?

Answer 11

If you write your polynomial in factored form, you can set each factor equal to 0 and solve for the value of \(x\) that makes that happen. Those values of \(x\) are the roots (or zeros as they are often called) of the polynomial.

Answer 12

Ohh okay that makes sense :)

Answer 13

Do you know what the Descartes rule of signs is?

Answer 14

Yes, I do.

Write the polynomial in descending order of exponents (like you have it).

Now going from left to right, count how many times the sign of the coefficients changes.

\[4x^3 + 4x^2 -8x = 0\]
We have 1 change of sign: +4 to -8

That means we have 1 positive, real root. If the number is 2 or greater, it may be reduced by a multiple of 2, but we have no way of knowing — we're just going to get the possible scenarios.

Next, we rewrite the polynomial, changing the signs of all of the terms with an odd exponent:
\[-4x^3+4x^2+8x=0\]Now repeat the counting process:
-4 to 4 is the only sign change, so we have 1 sign change which means 1 negative, real root. Again, if the number is 2 or greater, it may be reduced by a multiple of 2.

Finally, we look at the highest power of the variable in the polynomial to get the total number of roots. Any roots not spoken for by our counts of positive and negative real roots are pairs of complex roots in conjugate form: \(a\pm bi\) where \(i = \sqrt{-1}\). We must end up with a total number of roots that matches the highest power of the variable. Here we have 4 roots in total:

1 positive, real root,
1 negative, real root,
2 complex conjugate roots

If our counts had come out differently, we might have had
2 positive, real roots
2 complex conjugate roots
OR
0 positive, real roots
4 complex conjugate roots

(we wouldn't be able to distinguish between those two possibilities without some other knowledge).

We could also have had
2 negative, real roots
2 complex conjugate roots
OR
0 negative, real roots
4 complex conjugate roots

I'm not going to list the various combinations of real roots adding up to 4, because that's not the tricky part, it's this business of reducing the count by a multiple of 2 that takes a bit of getting used to.

Answer 15

Just one small correction: this equation has a total of 3 roots.

Answer 16

Yes, of course you are right about that! Kelsey did have another problem which had \(x^4\) and 3 terms but I can't deny that this one has \(x^3\). It was considerably past my usual hour for calling it a night, but I'm quite embarrassed by this!

Answer 17

@kelsey42 You're entitled to a full refund of everything you paid me for my help on this one due to my egregious mistake :-)