1. ## 3 homework question

1) Use long division to find the remainder when x^4-3x^2+5x-1 is divided by x^2-3

2) Solve the rational equation

x(2x+1) = 10 - 5
x - 2____ x-2_ 2

3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
Vertex is (3,-7)

2. Originally Posted by greenpumpkins
1) Use long division to find the remainder when x^4-3x^2+5x-1 is divided by x^2-3

2) Solve the rational equation

x(2x+1) = 10 - 5
x - 2____ x-2_ 2

3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
Vertex is (3,-7)
See here for help with polynomial long division. The quotient is $x^2+\frac{5x-1}{x^2-3}$. The remainder is $5x-1$

2) Solve the rational equation:

$\frac{x(2x+1)}{x-2}=\frac{10}{x-2}-\frac{5}{2}$

Multiply each term by the LCD of $2(x-2)$

$2(x-2)\left(\frac{x(2x+1)}{x-2}\right)=2(x-2)\left(\frac{10}{x-2}\right)-2(x-2)\left(\frac{5}{2}\right)$

$2x(2x+1)=2(10)-5(x-2)$

$4x^2+2x=20-5x+10$

$4x^2+7x-30=0$

$(x-2)(4x+15)=0$

$x=2 \ \ or \ \ x=-\frac{15}{4}$

But $x=2$ makes the denominator of the rational equation to be 0. So this solution is extraneous.

3) The line x = 3 is the axis of symmetry for the graph of a parabola. If the parabola contains the points (5, -3) and (-1,9), what is the equation for the parabola?
Vertex is (3,-7)

$y=a(x-h)^2+k$ Vertex form of the parabola with V(h,k)

Substitute one of the points on the parabola (5, -3) and the vertex (3, -7)

$-3=a(5-3)^2-7$

$-3=a(4)-7$

$a=1$

Therefore the equation for the parabola is

$\boxed{y=(x-3)^2-7}$ or $\boxed{y=x^2-6x+2}$

[tex]