# Using Descartes' Rule of Signs

#### Savior_Self

Use Descartes' rule of signs to determine the possible number of:
a) positive real roots of $$\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0$$
b) negative real roots of $$\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0$$

My answer to a) (zero positive real roots), was correct.

Zero positive real roots because there are zero sign changes..

and I had thought there would be zero negative roots as well. But it came back wrong.

Can someone explain to me why? Thanks.

#### masters

MHF Helper
Use Descartes' rule of signs to determine the possible number of:
a) positive real roots of $$\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0$$
b) negative real roots of $$\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0$$

My answer to a) (zero positive real roots), was correct.

Zero positive real roots because there are zero sign changes..

and I had thought there would be zero negative roots as well. But it came back wrong.

Can someone explain to me why? Thanks.
Hi Savior_Self,

$$\displaystyle f(-x)=4(-x)^3+2(-x)^2+7(-x)+9$$

$$\displaystyle f(-x)=-4x^3+2x^2-7x+9$$

Now, how many sign changes do you see?

• Savior_Self

#### spruancejr

To find the maximum number of positive real roots count the number of sign changes. To find the number of negative real roots replace all x values with a (-x) and perform the same operation of counting sign changes.

For example,

$$\displaystyle x^2+x+2$$ becomes...
$$\displaystyle (-x)^2+(-x)+2$$

• Savior_Self

#### ebaines

Using Descarte's method, to determine the number of negative roots you replace the x in the polynomial with -x, and see how many sign changes there are in the resulting equation:

$$\displaystyle 4(-x)^3 + 2(-x)^2 + 7(-x) + 9 =$$
$$\displaystyle -4x^3 + 2x^2 - 7x + 9$$

Note that there are 3 changes of sign as you read left to right. Thus there may be 3 negative roots, or possibly just 1 (since it's possible that 2 of the roots may actually be complex).

If you plot this function you'll see that there is one real negtive root. The two other roots are therefore complex.

• Savior_Self

#### Savior_Self

Awesome. Thanks guys. I didn't pick up on the x to (-x) step for some reason. I won't forget it now.

#### ebaines

There's another "trick" that you could have used. Remember that a polynomial of order N must have N roots. So since you already knew that there were zero positive roots, you know that there must be either (a) 3 negative roots, or (b) 1 negative root and 2 complex roots.