Consider the polynomial p (x) = x^3 + 3x^2 - 6x - 8. Calculate p (-4). Find all

Question

Consider the polynomial p (x) = x^3 + 3x^2 - 6x - 8. Calculate p (-4). Find all the roots of p. Write p as the product of three degree 1 polynomials: p (x) = (x - r_1) (x - r_2) (x - r_3). Give two distinct examples of degree 4 polynomials with zeros at 2, -4. and 4 (though they may also have zeros elsewhere). Let q (x) = x^2 - 4x + 3. Find the vertex and the zeros of q. Expand the expression q (3 + t) - q (3)/t to show that it is equal to p (t) = 2 + t. Use your results from 3a to make a plot of q on the interval [-1, 5]. On the same graph, make a plot of the line that has a slope of p (0) = 2 and includes the point (3, q (3)). Using long division or otherwise, find a polynomial G and a constant R such that r (x) = 12x^3 - 2x^2 - 24x + 2/6x - 1 = G (x) + R/6x - 1. By rounding R/599 = 0, use this result to approximate r (100) (without a calculator). What b the end behavior of the rational function f (x) = 4x^4 - 2x^2 + 5/8x^4 - 4x^2 + 6? Use long division to rewrite f (x) = G (x) + R/8x^4 - 4x^2 + 6, where R is a constant (the long division should finish in one step!). Do you see any connection between the result of the long division and the "end behavior' off?

Explanation / Answer

1]p(x) = x3 + 3x2 - 6x - 8

p(- 4) = (-4)3 + 3(-4)2 - 6(-4) - 8 = 0

b] p(x) = x3 + 3x2 - 6x - 8 = [x + 4][ax2 + bx + c]

since we know that x = - 4 is a factor of p(x)

dividing p(x) by (x + 4) we get:

ax2 + bx + c = x2 - x - 2

solving this gives x = - 1 and x = 2

therefore, the roots of p(x) are: x = - 4, - 1, 2

c] p(x) = (x + 4)(x + 1)(x - 2)

2] A four degree polynomial will be of the form: ax4 + bx3 + cx2 + dx + e

which will have 4 distinct solutions.

here, x = 2, - 4, 4

let x = 0 be the next root.

therefore, p(x) = x(x + 4)(x - 4)(x - 2) = (x2 - 2x)(x2 - 16) = x4 - 2x3 -16x2 + 32x + 0

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