CPhill

First, multiply the first two

(x - 1) ( x + 2)  =    x^2 - 1x + 2x - 2  = x^2 + x - 2

Now,  multiply   

( x^2 + x - 2) (x + 3)

Multiply the  three terms in the parenthses by each of the terms in the second set of parentheses

So we have

x^2 * x  +  x* x  -2*x  =   x^3 + x^2 - 2x       (1)

x^2 * 3 + x*3   -2*3  =  3x^2 + 3x - 6    (2)

Combine (1) and (2)   and we have

x^3 + x^2 + 3x^2 + 3x - 2x - 6  =

x^3 + 4x^2 +1x - 6

Here's the way to prove it, RP

AC and BC are the legs 

Show that AC = BC

Then....show that AC is perpendicular to AB

If you need help....let me know

We have the form

y = ax^3 + bx^2 + cx + d

Because f(0) = 0....then d = 0...and we can solve this system

a(-1)^3 + b(-1)^2 + c(-1) = 15

a(1)^3 + b(1)^2 + c(1) = 5

a(2)^3 + b(2)^2 + c(2) = 12          which translates to

-a + b  - c  =  15

a  + b  + c  =   5

8a + 4b + 2c =  12  

Adding the first  two equations produces  2b = 20 ⇒  b = 10

Using the last two  equations, we have

a + 10 + c = 5    ⇒  a + c =  - 5

8a + 40 + 2c = 12 ⇒  8a + 2c = -28 ⇒  4a + c = -14

Multiply the last equation by -1 and add to the frist equation

-4a - c = 14

a + c = 10

________

 -3a = 24   ⇒ a = -3

And a + c = -5 ⇒  c = -2   

So....the polynomial is

-3x^3 + 10x^2 - 2x    factor this and set to 0

x ( -3x^2 + 10x - 2) = 0

We  know that one root is  x = 0   [ this is one x intercept ]

So we need to solve this

-3x^2 + 10x - 2  = 0       multiply through by  -1

3x^2  - 10x + 2 = 0

Using the quadratic formula the other two x intercepts are

5 ±√19

______

     3

No prob....I  just answered the last one  a little while ago

By a property of medial triangles, each side of the medial triangle will be parallel to one side of its enclosing triangle....and the length of each of the sides of the medial triangle will be (1/2) the length of its respective parallel side....so....this must mean that the perimeter of triangle DEF will be twice that  of the perimeter of  triangle XYZ

And ....by like reasoning.....triangle ABC will have twice the perimeter of triangle DEF

So....the perimeter of ABC   =    5^(2)^2  =  20 units

"Flip" the fraction in the denominator and  multiply....and we have

2 + x              x 

____     *     ________   =

8x                 3 + x

(2 + x)        (x)

_____   *  _____ =

(3 + x)        8x

(2 + x)

______

8(3 + x)

Last one

f(x - 3)    shifts f(x) to the RIGHT by 3 units

2(x+3)            4x ( x + 2)

------------  *    ________           write as

 x ( x - 1)         10 (x + 3) 

2        4x          (x + 2)      (x + 3)

__ *   ___    *   _____ *  ______  =

10       x          (x - 1)       (x + 3)

1 * 4 * ( x + 2)

____________ =

 5 * (x - 1)

4 ( x + 2)

_______

5 (x - 1)

First of all...RICK doesn't form anything

I think it should be RKIC as shown below

Well anyway.....since two sides of a trapezoid must be parallel.....just show that the slopes between RK and KI and IC and CR are all unequal....and two of these must be equal to have a parallelogram

Since AB = 1

Then the side opposite angle A  = BC =  2   = the hypotenuse

And the side opposite angle B = AC  = sqrt(3)

 So....Let A = (0,0)    B = (1,0)    C = (0, sqrt(3) )

Call the midpoint of BC, T

And the coordinates of this point are   ( 1/2 , sqrt(3)/2)

So....AT is a median of triangle ABC

And centroid G will be (2/3) of the distance from A to T

And this distance is  (2/3)sqrt [ (1/2)^2 + (sqrt(3)/2)^2 ] = (2/3) sqrt [ 1/4 + 3/4] =

(2/3) sqrt (1) =     (2/3)  = AG