Question

Can somebody help me with the 3 questions?

Answer 1

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: $x = -2$

Complex zeros: $x_{2,3}=-3\pm 2i$

Step-by-step explanation:

1. Given equation of parabola

$s(t)=-16t^2+88t+48$

a) The rocket reaches its maximum height at the vertex of parabola. Find t-coordinate of the vertex:

$t_v=\dfrac{-b}{2a}\\ \\=\dfrac{-88}{2\cdot (-16)}\\ \\=\dfrac{11}{4}\\ \\=2.75\ seconds$

b) The maximum height is s-coordinate of the vertex. Find it:

$s\left(\dfrac{11}{4}\right)\\ \\=-16\cdot \left(\dfrac{11}{4}\right)^2+88\cdot\left(\dfrac{11}{4}\right)+48\\ \\=-121+22\cdot 11+48\\ \\=169\ feet$

2. For x – 3 to be a factor of  $f(x)=6x^4-11x^3-35x^2+34x+24,$ the Factor Theorem says that x = 3 must be a zero of  f(x). Check it (whether f(3)=0):

$f(3)\\ \\=6\cdot 3^4-11\cdot 3^3-35\cdot 3^2+34\cdot 3+24\\ \\=6\cdot 81-11\cdot 27-35\cdot 9+102+24\\ \\=486-297-315+126\\ \\=0$

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:

$f(x)\\ \\=6x^4-11x^3-35x^2+34x+24\\ \\=6x^4-18x^3+7x^3-21x^2-14x^2+42x-8x+24\\ \\=6x^3(x-3)+7x^2(x-3)-14x(x-3)-8(x-3)\\ \\=(x-3)(6x^3+7x^2-14x-8)\\ \\=(x-3)(6x^3+12x^2-5x^2-10x-4x-8)\\ \\=(x-3)(6x^2(x+2)-5x(x+2)-4(x+2))=\\ \\=(x-3)(x+2)(6x^2-5x-4)\\ \\=(x-3)(x+2)(6x^2+3x-8x-4)\\ \\=(x-3)(x+2)(3x(2x+1)-4(2x+1))\\ \\=(x-3)(x+2)(2x+1)(3x-4)$

3. $x=-2$ is a zero of the function $f(x)=x^3+8x^2+25x+26,$ then

$f(x)\\ \\=x^3+8x^2+25x+26\\ \\=x^3+2x^2+6x^2+12x+13x+26\\ \\=x^2(x+2)+6x(x+2)+13(x+2)\\ \\=(x+2)(x^2+6x+13)$

Find the discriminant of the quadratic polynomial $x^2+6x+13;$

$D=6^2-4\cdot 1\cdot 13=36-52=-16$

This expression has no more real zeros (the discriminant is less than 0), it has two complex zeros:

$x_{1,2}=\dfrac{-6\pm \sqrt{-16}}{2\cdot 1}=\dfrac{-6\pm 4i}{2}=-3\pm 2i$