question below?

Question

1. Given the polynomial P(x) = 2x3 - x2 - 13x – 6 and one of its factors: x + 2.

a. How many zeros will there be? How do you know that?

b. Find the remaining factors of the polynomial.

c. Please explain in words how you found the others

2. Analyze the roots of the polynomial P(x) = x4 + 3x3 - 5x2 + 2x – 4

a. Tell how many possible positive rational roots.

b. Tell how many possible negative rational roots.

c. Tell how many possible complex roots.

Answer 1

a: There will be 3 zeros or roots
b:The remaining factors are (2x+1)(x-3)
C:
P(x) = 2x3 - x2 - 13x – 6 and x+2 is a factor

Step 1: you divide P(x) by x+2

2x3 - x2 - 13x – 6 /(x+2)
= 2x2 - 13x – 6

You factor 2x2 - 13x – 6 =(2x+1)(x-3)

P(x) = 2x3 - x2 - 13x – 6= ( x+2)(2x+1)(x-3)

2
P(x) = x4 + 3x3 - 5x2 + 2x – 4
You need a graph and I do not know how to show it on this sheet.

Answer 2

1. a. A cubic may have one, two, or three real roots. Without finding the others, we can tell that this one will have three real roots because its first coefficient is positive, its constant term is negative, and its given root is negative.

An odd-degree polynomial with positive first coefficient is below the x-axis when x is large and negative, and above the x-axis when x is large and positive. In the case of a cubic, it may cross the x-axis once (one real root and two complex), cross once and be tangent once (two real roots, one of them of order two), or cross three times (three real roots). So if it had only one real root, it wouldn't get back down to the x-axis after crossing it once, and if it had two it wouldn't get below.

b. The remaining factors are x - 3 and 2x + 1

c. I tried plugging x = 3 into P(x), because it's one of the possibilities to make the constant term come out to 6. Then I needed a factor of 2 for the cubic term and no more factors for the constant term, so I expanded (x - 3)(2x + 1)(x + 2) to verify that it's P(x).

2. a&b. A quartic polynomial can have up to four roots. Because the constant term is negative, they can't all be on the same side of it, since an even-degree polynomial is positive both as x goes toward infinity and as it goes toward negative infinity. The possible rational roots are plus or minus 1, 2, and 4. But the factors have to multiply together to give -4 as the constant term, and if you had all three it would be a multiple of eight. So there are at most two positive rational roots. And by the same reasoning there are at most two negative rational roots.

c. There are at most two complex roots: there's at least one positive and one negative real root because P(0) is negative and P is of even degree, which leaves at most two complex ones out of the four roots of a quartic.

P.S. The only way to show a graph here, akaik, is to put it on an image-hosting site and just post the link.

Answer 3

P(x) has only one sign change so there can be only one positive root. P(-x) = -2x^3-x^2+13x-6 has two sign changes so there are either 2 or 0 negative roots.
Since x=-2 is given as a root there must be two negative roots and one positive root. This is DeCartes's Rule of Signs.

long division gives 2x^2 -5x -3 = (2x+1)(x-3)
So x = -1/2 or x = 3 are the additional zeroes.

P(x) has 3 sign changes so there are either 3 positive roots or 1 positive root. P(-x) = x^4 -3x^3 -5x^2 -2x -4 has 1 sign change so it has 1 negative root. So there could be be two imaginary complex roots.