Polynomial Function when C is a Constant

[kendang]

Question:

Consider the polynomial function f(x)=(x-3)(x-1)(x+2)^2+C where C is a constant. Sketch this polynomial function when C=0. For each of the following questions below, fully explain your mathematical reasoning. Hint: carefully look at your sketch of f(x) when C=0.

a) Determine the value of C for the function f(x) to have four different real roots.
b) Determine the value of C for the function f(x) to have three real roots (one root must have a double of course!)
c) Determine the value of C to have at most two real roots.

So I guess when C=0, the graph looks like this:

a) The graph when C=0 only has 3 roots, so I need to shift the graph down by making the value of "C =< -1". Is that how you express it mathematically?

b) When C=0, there are only 3 roots. What are real roots and what are double roots?

c) The graph when C=0 only has 3 roots, so I need to shift the graph up by making the value of "C => -1". Is that how you express it mathematically?

 

[Deleted member 4993]

Question:

Consider the polynomial function f(x)=(x-3)(x-1)(x+2)^2+C where C is a constant. Sketch this polynomial function when C=0. For each of the following questions below, fully explain your mathematical reasoning. Hint: carefully look at your sketch of f(x) when C=0.

a) Determine the value of C for the function f(x) to have four different real roots.
b) Determine the value of C for the function f(x) to have three real roots (one root must have a double of course!)
c) Determine the value of C to have at most two real roots.

So I guess when C=0, the graph looks like this:

a) The graph when C=0 only has 3 roots, so I need to shift the graph down by making the value of "C =< -1".

What happens when C = -0.25 or C = -0.5 or C = -0.75

Is that how you express it mathematically?

b) When C=0, there are only 3 roots. What are real roots and what are double roots? Look in your textbook or do a google search for those definitions

c) The graph when C=0 only has 3 roots, so I need to shift the graph up by making the value of "C => -1".

What happens when C = 0.25 or C = 0.5 or C = 0.75



Is that how you express it mathematically?

 

[soroban]

Hello, kendang!

Consider the polynomial function: \(\displaystyle f(x)\:=\x-3)(x-1)(x+2)^2+C\) where \(\displaystyle C\) is a constant.
Sketch this polynomial function when \(\displaystyle C=0.\)

So I guess when \(\displaystyle C=0\), the graph looks like this:

For each of the following questions below, fully explain your mathematical reasoning.
Hint: carefully look at your sketch of f(x) when \(\displaystyle C=0.\)

a) Determine the value of \(\displaystyle C\) for the function to have four different real roots.

The graph when \(\displaystyle C=0\) has only 3 roots, so I need to shift the graph down . Right!
. . by making the value of \(\displaystyle C \le -1\) . Wrong!
Is that how you express it mathematically?

We want \(\displaystyle C\) to be any negative number: .\(\displaystyle C\,<\,0.\)

b) When \(\displaystyle C=0\), there are only 3 roots.
What are the real roots and what is the double root?

Look at your graph.
Two of the roots are 1 and 3.
The double root is -2.

c) The graph when \(\displaystyle C=0\) has only 3 roots, so I need to shift the graph up . Right!
. . by making the value of "C => -1". . Wrong!
Is that how you express it mathematically?

We want \(\displaystyle C\) to be any positive number: .\(\displaystyle C\,>\,0.\)

 

[JeffM]

Question:

What are real roots and what are double roots?

Are you aware of the distinction between real and complex numbers?

There is a theorem of algebra that says that any polynomial of degree n has up to n distinct complex roots. This can also be expressed as saying that any polynomial of degree n has n complex roots, but that some may be duplicates. Let's look at this in the context of three very simple quadratics.

Example I: x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 17 = 0. There are no real roots to this quadratic as you can quickly see if you graph it. (You can also prove it algebraically.)
There are, however, two complex roots, namely x = - 4 + i and x = - 4 - i. This is easily proved.
Step 1: (x + 4 - i) * (x + 4 + i) = x[sup:3c3dpswd]2[/sup:3c3dpswd] + 4x + ix + 4x + 16 + 4i - ix - 4i - i[sup:3c3dpswd]2[/sup:3c3dpswd] = x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 16 - (- 1) = x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 17.
Step 2: If x = - 4 + i, then x + 4 - i = 0 and so x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 17 = 0.
Step 3: If x = - 4 - i, then x + 4 + i = 0 and so x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 17 = 0.
So here is an example of a polynomial with no real roots. (All such polynomials are of even degree.) But it does have 2 complex roots.

Example II: x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 16 = 0. There is one real root at x = - 4. Let's prove this.
Step 1: (x + 4) * (x + 4) = x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 16.
Step 2: (x + 4) = 0 if x = - 4 so x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 16 = 0 if x = - 4.
Step 3: (x + 4) * (x + 4) = (x + 4)[sup:3c3dpswd]2[/sup:3c3dpswd] so at any other value of x it must be true that (x + 4)[sup:3c3dpswd]2[/sup:3c3dpswd] = x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 16 > 0.
Notice however that (x + 4) * (x + 4) has two terms so it is reasonable to say that it the quadratic does have two real roots that happen to be identical.
But any real number is also a complex number. That is, - 4 = - 4 + (0 * i). So this quadratic also has two complex roots (although they are not distinct).

Example III: x[sup:3c3dpswd]2[/sup:3c3dpswd] + 8x + 15 = 0. As you can easily satisfy yourself, this has two real roots, one at - 3 and one at - 5. But all real numbers are also complex numbers. So again we have 2 complex roots.

Every quadratic has two complex roots, which may be duplicates.
Every cubic has three complex roots, at least one of which must be real.
And so on.

Does this answer your question?