Answer:
1.
a. 2.75 seconds
b. 169 feet
2. x - 3 is a factor
Other factors: x + 2, 2x + 1, 3x - 4
3. Real zeros: [tex]x = -2[/tex]
Complex zeros: [tex]x_{2,3}=-3\pm 2i[/tex]
Step-by-step explanation:
1. Given equation of parabola
[tex]s(t)=-16t^2+88t+48[/tex]
a) The rocket reaches its maximum height at the vertex of parabola. Find t-coordinate of the vertex:
[tex]t_v=\dfrac{-b}{2a}\\ \\=\dfrac{-88}{2\cdot (-16)}\\ \\=\dfrac{11}{4}\\ \\=2.75\ seconds[/tex]
b) The maximum height is s-coordinate of the vertex. Find it:
[tex]s\left(\dfrac{11}{4}\right)\\ \\=-16\cdot \left(\dfrac{11}{4}\right)^2+88\cdot\left(\dfrac{11}{4}\right)+48\\ \\=-121+22\cdot 11+48\\ \\=169\ feet[/tex]
2. For x – 3 to be a factor of [tex]f(x)=6x^4-11x^3-35x^2+34x+24,[/tex] the Factor Theorem says that x = 3 must be a zero of f(x). Check it (whether f(3)=0):
[tex]f(3)\\ \\=6\cdot 3^4-11\cdot 3^3-35\cdot 3^2+34\cdot 3+24\\ \\=6\cdot 81-11\cdot 27-35\cdot 9+102+24\\ \\=486-297-315+126\\ \\=0[/tex]
So, x = 3 is zero of the function f(x) and x - 3 is the factor of the function f(x). Rewrite the function as follows:
[tex]f(x)\\ \\=6x^4-11x^3-35x^2+34x+24\\ \\=6x^4-18x^3+7x^3-21x^2-14x^2+42x-8x+24\\ \\=6x^3(x-3)+7x^2(x-3)-14x(x-3)-8(x-3)\\ \\=(x-3)(6x^3+7x^2-14x-8)\\ \\=(x-3)(6x^3+12x^2-5x^2-10x-4x-8)\\ \\=(x-3)(6x^2(x+2)-5x(x+2)-4(x+2))=\\ \\=(x-3)(x+2)(6x^2-5x-4)\\ \\=(x-3)(x+2)(6x^2+3x-8x-4)\\ \\=(x-3)(x+2)(3x(2x+1)-4(2x+1))\\ \\=(x-3)(x+2)(2x+1)(3x-4)[/tex]
3. [tex]x=-2[/tex] is a zero of the function [tex]f(x)=x^3+8x^2+25x+26,[/tex] then
[tex]f(x)\\ \\=x^3+8x^2+25x+26\\ \\=x^3+2x^2+6x^2+12x+13x+26\\ \\=x^2(x+2)+6x(x+2)+13(x+2)\\ \\=(x+2)(x^2+6x+13)[/tex]
Find the discriminant of the quadratic polynomial [tex]x^2+6x+13;[/tex]
[tex]D=6^2-4\cdot 1\cdot 13=36-52=-16[/tex]
This expression has no more real zeros (the discriminant is less than 0), it has two complex zeros:
[tex]x_{1,2}=\dfrac{-6\pm \sqrt{-16}}{2\cdot 1}=\dfrac{-6\pm 4i}{2}=-3\pm 2i[/tex]
