Question

State how many imaginary and real zeros the function has.

f(x) = x^3 + 5x^2 + x + 5

Answer 1

An nth degree polynomial has n total real and imaginary roots (counting repeated roots).

Answer 2

Can you explain how that applies to this equation? I don't really understand.

Answer 3

Do you know how to tell what the degree of a polynomial is?

If I ask you, "what is the degree of \(x^3 + 5x^2 + x + 5\)," do you know the answer?

Answer 4

no

Answer 5

The degree of the polynomial is the degree of the leading term.

The degree of a term is the total of all the powers (exponents) in that term.

So, e.g.:

\(\Large f(x)=x\) is a 1st degree polynomial (or "has degree of 1")
\(\Large f(x)=5x^5-2x^2+1\) has degree 5
\(\Large f(x)=-7x^{11}-x\) has degree 11

Make sense?

Answer 6

So what's the degree of \(\Large x^3 + 5x^2 + x + 5\) ?

Answer 7

3. so for the question I gave it would be the 3 real and 3 imaginary?

Answer 8

Not quite. Yes, the degree is 3.

An nth degree polynomial has n total real and imaginary roots (counting repeated roots).

The TOTAL roots is 3. Can be 3 real, or 1 real and 2 imaginary (we won't go into why it can't be 3 imaginary here - not necessary for this question).

Answer 9

my answer choices say:
0 imaginary; 3 real

1 imaginary; 2 real

3 imaginary; 0 real

2 imaginary; 1 real
it has both of your choices so I'm confused.

Answer 10

f(x) = x^3 + 5x^2 + x + 5
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As stated above, expect three roots for this cubic polynomial.

f(x) = x^3 + 5x^2 + x + 5 factors into((x+5) (x² +1)

Setting ((x+5) (x² +1) = 0,
x + 5 =0 or x² +1 = 0

Solve each of those for x, and the nature of the roots will reveal themselves.

@faariat

Answer 11

Ahhh, sorry... I misinterpreted the question, did not realize that it wanted you to pin down the types of roots.

As @Directrix demonstrates above, if you can factor the polynomial, you can determine the types of factors that way. The factor x+5 has one real zero, so the only question is, what kind of roots does x² +1=0 have? It is a quadratic, so has to have either 2 real or 2 imaginary roots.

Some of this analysis depends on what you have been showed in class, and what the teacher expects you to do, also. Another method, if you're allowed to use it, is to simply graph the function on a graphing calculator or other technology and look at the graph. Each REAL root corresponds to an x-intercept. If you had, for example, 1 x-intercept, then there are only two possibilities:
1. You have one real, and two complex, roots. or...
2. You have a function that is a perfect cube, e.g., something that factors to: \(y=(x-c)^3\). But that is clearly not the case here, since the last term of the function is not a perfect cube.

Answer 12

By the way, in that graphing analysis, if you had THREE x-intercepts, obviously that would mean 3 real roots (each real root is an x-intercept on the graph; complex roots do NOT produce x-intercepts).

If you had TWO x-intercepts, that would still mean THREE real roots, but one of them is a "repeated root", meaning that one of the factors of the polynomial is a perfect square with a real root.