Question

MAT 171

1. The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=-4
Find a possible formula for P(x)


2. The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=4 and roots of multiplicity 1 at x=0 and x=-4. It goes through the point (5, 36).
Find a formula for P(x).


3. The polynomial of degree 3. P(x), has a root of multiplicity 2 at x=3 and a root of multiplicity 1 at x=-2. The y-intercept is y=-1.8.
Find a formula for P(x).

Answer 1

Using the Factor Theorem, the polynomials are given as follows:

1. $P(x) = x^5 + 2x^4 - 7x^3 + x^2$

2. $P(x) = 0.8(x^4 - 4x^3 - 16x^2 + 64x)$

3. P(x) = -0.1(x³ - 4x² - 3x + 18)

What is the Factor Theorem?

The Factor Theorem states that a polynomial function with roots $x_1, x_2, \codts, x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)$

In which a is the leading coefficient.

Item a:

The parameters are:

$a = 1, x_1 = x_2 = 1, x_3 = x_4 = 0, x_5 = -4$

Hence the equation is:

P(x) = (x - 1)²x²(x + 4)

P(x) = (x² - 2x + 1)(x + 4)x²

P(x) = (x³ + 2x² - 7x + 1)x²

$P(x) = x^5 + 2x^4 - 7x^3 + x^2$

Item b:

The roots are:

$x_1 = x_2 = 4, x_3 = 0, x_4 = -4$

Hence:

P(x) = a(x - 4)²x(x + 4)

P(x) = a(x² - 16)x(x - 4)

P(x) = a(x³ - 16x)(x - 4)

$P(x) = a(x^4 - 4x^3 - 16x^2 + 64x)$

It passes through the point x = 5, P(x) = 36, hence:

45a = 36.

a = 4/5

a = 0.8

Hence:

$P(x) = 0.8(x^4 - 4x^3 - 16x^2 + 64x)$

Item 3:

The roots are:

$x_1 = x_2 = 3, x_3 = -2$

Hence:

P(x) = a(x - 3)²(x + 2)

P(x) = a(x² - 6x + 9)(x + 2)

P(x) = a(x³ - 4x² - 3x + 18)

For the y-intercept, x = 0, y = -1.8, hence:

18a = -1.8 -> a = -0.1

Thus the function is:

P(x) = -0.1(x³ - 4x² - 3x + 18)

More can be learned about the Factor Theorem at https://brainly.com/question/24380382

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