CPhill

x^4+2x^3−16x^2−2x+15

Note  that we can use a special  "trick" to write this in a slightly different manner

This will make the polynomial easily "factorable"

[ P.S.  - This does not always work...but it just happens to, here  ]

(x^4 + 2x^3 - 15x^2) - ( x^2 + 2x -15)    factor

x^2 ( x^2 + 2x - 15) - 1 ( x^2 + 2x - 15)   

(x^2 - 1) ( x^2 + 2x - 15)

(x + 1) ( x - 1) ( x + 5) ( x - 3)

Setting each factor to 0  and solving for x  gives the roots  (zeros)

x =  - 1 ,  x  = 1  ,   x = -5     x  =  3

f(x)=x^3+5x^2+6x

Let's examine each statement

First one

When  x  = 2      f(2)  =  (2)^3 + 5(2)^2 + 6(2)   =  8 + 20 + 12  =  40

So....this one is false

Second one

x^3 + 5x^2 + 6x     factor

x( x^2 + 5x + 6)

x(x + 3) ( x + 2)     

So....this one is true

Third one

This is false...it will only have a remainder of 0  if divided by one of its factors....and (x - 2) is not a factor

Last one

f(-2)   =  0  is true.....since  ( x + 2)  is a factor, - 2  is a root  [ it will produce a 0  when plugged into the function  and evaluated  ]

Very nice, EP.....I didn't know you were so well-versed in Chemistry!!!

[ I used to be able to do this stuff in High School, but....I've slept since then......]

Well...to be honest....the first time I encountered this one it kind of "threw" me, too.  But, sometimes, just one little "trick"  is the key to solving a problem.....in this case..the "similar triangles" thing...

I think this might be similar to what you need :

Note that  triangles  BHC  and BCA  are similar...so...

HC / BC   = CA / BA     so

h / a  =  b / ( x + y)       (1)

Working with your expansion...note that

x^2 + h^2  = a^2      and

y^2 + h^2  = b^2

So...subtracting out the equal parts we are left with

2xh  + 2yh   =  2ab        (3)    divide through by 2

xh + yh  = ab      factor the left side

h ( x + y)  =  ab      which we can rearrange as

h / a   =  b / ( x+ y)      (2)

And since this is  equal to (1)   and  it's just a rearrangement of (3)  adding all the other equal parts of the expansion shows that

(x^2 + h^2)  + (y^2 + h^2) + ( 2xh + 2yh) =  a^2  + b^2 + 2ab

( x^2 + 2xh + h^2) + ( y^2 + 2yh + h^2)  =  ( a + b)^2

( x + h)^2   +  ( y + h)^2  =  (a + b)^2      ....!!!!

Don't know if this is what you are looking for....but....here's a solution using Calculus

The first  function  can be written  as

y  =  (1/√2) ( 2x^2 - 2x + 1)^(1/2)

The derivative of this function is

y'  =   ( 1/ sqrt(8)) ( 2x^2 - 2x +1)^(-1/2) ( 4x - 2)

We only need to solve this to find a possible   x value of a max / min  of this function [a critical point ]

4x - 2    =  0

4x  = 2

x = .5  =  1/2

And the y value  at this point  =    [1/ √2]  ( 2(1/2)^2 - 2(1/2) + 1)^(1/2)  =

[1/√2 ] ( 1/2 - 1 + 1 )^(1/2)   =   (1/4)^(1/2)  =  1/2

This point is a min  on [ 0, 1]  because

And  at x  = 0  the function value  = 1/√2  ≈ .707

And at x  = 1 the function value  =  1/√2 ≈ .707

So...the point  (.5, .5)  is the lowest point on this graph  from (0, 1)

The second  function can be written as

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

The derivative of this is

y '  =  (x^2 + 1)^(-1) -  x (x^2 + 1)^(-2) * 2x       factor this

y '  =   (x^2 + 1)^(-2)   [ (x^2 + 1 - 2x^2 ]

y'  =  ( x^2 + 1)^(-2) [ 1 - x^2)

To  find  the possible x  that minimizes / maximizes this function on the interval  we can solve this

1 - x^2  =  0

(1 - x) ( 1 + x)  = 0     set both factors to 0  and solve for x  and we get that

 x = 1   or  x  = -1       and these are the critical points of this function

We can ignore the second answer since it's out of the interval

So...the function has either a  max or min value  at x  = 1

At x  = 1  the y value  is     1 / [ 1^2 + 1] =  1/2

At x  = 0  the y value is  0 / [ 0^2 + 1] =  0

Since the value a x = 1  is greater than at x = 0....this fuction has a  max value on [ 0, 1]  of  1/2....and at all other points in (0,1)   y is less  than 1/2

But  the first function has a  minimum y value of 1/2  on (0,1)

So..on the interval  (0,1) the first function is always greater than the second function

So

√[ (2x^2 - 2x + 1) / 2 ]  ≥   1 /  [  x + 1/x ]     on  (0, 1)

Monthly interest rate  =  .035 / 12  = 7/2400

Five years

n  = 60

33000 (7/2400)   ( 1 + 7/2400)^ 60  /  [ ( 1 + 7/2400)^60 - 1 ]  ≈ $600.33 / mo

Six years

n = 72

33000 (7/2400)   ( 1 + 7/2400)^ 72  /  [ ( 1 + 7/2400)^72 - 1 ]  ≈ $508.81 / mo

Refresh my memory, NSS....isn't the "Rule of Housing"  25%  of the annual salary / 12

If so.....we have

42000 * .25  /  12   =    $875 / mo

Don't know much about this, NSS...but...here's my best attempt...

a  =  57.50  + 49.50 + 18.20 + 38.90 + 85.60  + 2.00 + 20.00  =  $271.70

b  = 59.10  + 49.50   + 18.20 +  42.30 + 90.80 + 2.00 + 20.00 = $281.90

c = $271.70

d =  281.90 + 76.10   = $358

e  =  c + d   =   271.70 + 358   = $ 629.70   

P  =  10000  ( .05/12) ( 1 + .05/12)^(36)  / [ ( 1 + .05/12) - 1 ] =

10000 (1/240)  ( 1 + 1/240)^36 / [ ( 1 + 1/240)^36 - 1]  =

(10000 /240) ( 1 + 1/240)^36 / [  ( 1 + 1/240)^36 - 1  ]  ≈

$299.71