how do i determine the number of real zeros of the function

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

X^3 + 4 X^2 + X -6

Sum of the roots has to be : X1 + X2 + X3 = -4

X1 . X2. X3 = -6 write down all factors of 6 : 1,2, 3

we see that -3 - 2 +1 =-4

and (-3)( -2) (1) = -6

Therefore the 3 rational roots are: X1 = -3 X2 = -2 X3=1

Polynomial can be factored as: X^3 + 4 X^2 + X - 6 = ( X+3) ( X+2) ( X-1)

This is what you do without calculation:

Michael F. | Mathematics TutorMathematics Tutor

Since the function is a polynomial with real coefficients, and is of odd degree, it must have at least one real root.

Using the rational roots theorem all the roots must be plus or minus factors of 6. As you can see 1 is such a root.

Using synthetic division

1|1 4 1 -6

1 5 6

1 5 6 0 that is x^{2}+5x+6. So the polynomial is (x-1)(x^{2}+5x+6) IF ALL YOU ARE AFTER IS HOW MANY REAL ROOTS then

5^{2}-4×6=1, the discriminant is positive, the other roots are also real.

You can use factor theorem and synthetic division to answer the question.

f(x) = x^{3} + 4x^{2} + x - 6

Since f(1) = 0, by factor theorem x-1 is a factor.

Using synthetic division,

1 | 1 4 1 -6

........1 5 6

-----------------

....1 5 6 | 0

So, f(x) = (x-1)(x^{2}+5x+6) = (x-1)(x+2)(x+3)

Answer: There are three real zeros: x = 1, -2, -3.

Dear Queneshia,

I hope I'm interpreting this correctly:

f(x) = x^{3} + 4x^{2} + x - 6 = 0

The expression for f(x) cannot, unfortunately, be factored.

One solution that suggests itself immediately is x = 1

F(1) = (1)^{3} + 4*(1)^{2} +1 = 6 (which is true)

Are you allowed to graph the function, Queneshia? Sometimes for expressions that cannot be factored, graphing might be your only option.

Let's make a table of x and f(x) values to graph:

x f(x)

-4 -10

-3 0

-2 0

-1 -4

0 -6

1 0

2 20

Once you graph the function, you can easily see that the zeros are -3, -2 and 1. There are no others.

Question:

X1 + X2 + X3 = -b/a = -3 - 2 + 1 = -4 ( True)

Then :

X1 . X2 . X3 = d/a = -6, however (1) ( -3) ( -2) = +6 , doesn't seem to be consistent.

Parviz: (x-a)(x-b)(x-c)=x^{3}-(a+b+c)x^{2}+(ab+ac+bc)x-abc

In our case -(1)(-2)(-3)=-6

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