Math help?

Question

Use Descartes' Rule of Signs and the Rational Zeros Theorem to find all real zeros, then use the zeros to factor over the real numbers.

f(x) = x^4 + 6x^3 + 11x^2 + 12x + 18

Answer 1

Hi,

Descartes' Rule of signs would tell you that since there are no sign changes in the equation, then there are no positive real roots. Possible negative rational roots are -1, -2, -3, -6, -9, or -18. -3 is a double root, which is shown by synthetic division below.
....____________
-3)1..6..11..12..18
....__-3_-9_-6_-18_
.....1..3....2...6.....0 <== This remainder shows -3 is a root.

Now take the coefficients in front of the remainder and divide by -3 again.

...._________
.-3)1..3...2...6
....__-3_.0_-6_
......1..0..2....0 <== This remainder shows -3 is a root again.

The expression left is x² + 2 = 0 Solving this:
x² = -2
√x² = ±i√2
x = ±i√2

So the 4 roots are -3, -3, and ±i√2

I hope that helps!! :-)

Answer 2

Descartes' rule of signs states that the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or less than it by a multiple of 2.

Looking at your terms in order, they are all positive, so there are no sign changes (thus no positive roots).

The corollary to the rule (for negative roots) is switch the sign on the odd powers. Check the number of sign changes and that (or two less than it) will give the number of negative roots.

In your example, when you negate the odd power terms you get:
x^4 - 6x^3 + 11x^2 - 12x + 18.

So there will be 4 or 2 negative roots.

Roots will be factors of the last coefficient (18) divided by factors of the first coefficient (1). In other words, try -1, -2, -3, -6, -9 and -18.

A little trial and error shows that -3 is a root, so factor out an (x + 3) term using synthetic division.

(x + 3)(x^3 + 3x^2 + 2x + 6)

Now try the following roots -1, -2, -3, -6. Again we find that x = -3 is another root, a double root essentially.

So use synthetic division again to factor out another (x + 3) term:
(x + 3)(x + 3)(x² + 2)

You can now use the quadratic formula or just inspection to see that there will be no real root for the last part. x² = -2, so x = sqrt(-2). That means x = ±i√2

Your real roots are -3 and -3 (double root).
Your imaginary roots are i√2 and -i√2.

I graphed the function just to confirm that there are only two real roots (a double root at -3, where it just touches the x-axis, rather than crossing). See the graph below.

Answer 3

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Answer 4

so sorry,I haven't been taught abt those rule.I can't help you.

Answer 5

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Answer 6

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