CPhill

(x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5

(x^2 + ax + b) ( x^2 + cx + d)  =

    x^4 + cx^3 + dx^2

+        +ax^3  + acx^2 + adx

+                   +  bx^2 +  bcx   +  bd

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   x^4  + (a + c)x^3  + ( ac + b + d)x^2 + (bc + ad)x  + bd

Equating coefficients we have this system of equations

 a + c  =  1

ac + b + d  = -2

ad + bc  =  17

bd  = - 5

Since  bd  = - 5.....then we have the following possibilities for  ( b, d)  = (-1, 5), ( 5, -1) , ( 1, -5) or ( -5, 1)  

Check   (b, d)  = ( -1, 5)

ad + bc  = 17  ⇒   a(5) + ( -1)c  =17   ⇒  5a - c  = 17     (1)

a + c  = 1   (2)

Add (1) and (2)  and we get that  6a = 18b⇒  a  = 3

And  using (1), 5(3) - c  = 17 ⇒  -c = 2  ⇒ c  =-2

So...in this  situation   a  = 3, b = -1 , c = -2 , d = 5     and their sum =  5

Check  (b,d)  = (5, -1)

a(-1) + (5)c  = 17  ⇒  -a + 5c  =  17       (3)

a + c  =  1   (4)

Add (3) and (4)  and we get that 

6c  = 18  ⇒ c  = 3     

And using (3),  -a + 5(3)  = 17  ⇒  -a + 15  = 17 ⇒  -a  = 2 ⇒  a  =-2

So....in this situation   a = -2, b = 5, c = 3   d  = -1    and the sum is still = 5

Check for yourself  that  (b,d)  = (1, - 5)  and ( -5, 1)  don't provide integer solutions

Check   

(x^2+3x-1)(x^2-2x+5) = x^4+x^3-2x^2+17x-5

(x^2 -2x + 5) ( x^2 + 3x -1)  = x^4+x^3-2x^2+17x-5

Here you go....we need this :

[ x - 3 ]  [ x - (2 + 2i) ] [ x - (2 - 2i) ]  =  0

[x - 3 ]  [ x^2 - (2 + 2i)x - (2 - 2i)x +  (2 + 2i)(2-2i) ] = 0

[ x - 3 ]  [ x^2 -2x - 2ix - 2x + 2ix  + 4 - 4i^2 ]   =  0

[ x - 3 ]  [ x^2 - 4x + 4 + 4 ]  = 0

[x - 3 ] [ x^2 - 4x + 8 ]  = 0

[ x^3 - 4x^2 + 8x - 3x^2 + 12x - 24 ] = 0

x^3 - 7x^2 + 20x - 24  =  0

9. Given that x is real  and  x^3  + 1/x^3  =  52  , find x + 1/x

(x + 1/x)^3  =  x^3   + 3x^2(1/x) + 3x(1/x)^2 + (1/x)^3      simplify

(x + 1/x)^2  = x^3 + 3x + 3/x  + 1/x^3

(x + 1/x)^2  =  [ x^3 + 1/x^3 ]  + 3 [ x + 1/x]

(x + 1/x)^3  =  52  + 3[ x + 1/x]

(x + 1/x)^3   - 3 (x + 1/x) - 52  = 0           let   (x + 1/x)   =   a      ......so we have

a^3  - 3a  -  52  =  0

By the Rational Roots Theorem, the possible roots are  ±( 1, 2, 4, 13, 26, 52 )

Note that   4    produces a solution because

4^3  - 3(4)  - 52  =

64  - 12  - 52   =  

64 - 64   = 0

Therefore    a   =   (x + 1/x)   =  4

Thanks, heureka.....

Here is the graph  of the max rectangle :

(x + 4) ( x + 2) (2x+ 1) (x  - 2) = 0

(x^2 - 4) ( x + 4) (2x + 1)  = 0

(x^2 - 4) (2x^2 + 9x + 4)  =  0

2x^4 + 9x^3 + 4x^2

                        -8x^2   - 36x   - 16     

__________________________

2x^4 + 9x^3 - 4x^2 - 36x -  16   = 0 

The error was made in multiplying (x + 4) (2x + 1). The first term is 2x^2, not x^2.

Your answer is as good as mine, Tom.....the change-of-base works well, too.....

On our calculator, type 

log (5, 3)     and hit  the  " = " key

This is the evaluation of  log ₃ 5

Widgets produced on Monday  =  widgets per hr   * hrs worked  = wt   =  (2t) * t  = 2t^2 

Widgets produced on Tuesday  =  (w + 4) *  (t -2)  =  wt + 4t - 2w - 8  = 2t^2  + 4t - 2(2t) - 8  =  2t^2  - 8

So  he produces 8 more widgets on Monday than Tuesday

c  = √ [a^2 + b^2]     square both sides

c^2  =  a^2 + b^2     subtract b^2 from both sides

c^2 - b^2  =  a^2       take the positive root of both sides

√[c^2 - b^2 ] = √a^2

√[c^2 - b^2 ]  = a

Note  that   at x = -1,  y = 0   so  x = -1  is a root

And note that when x = 0, y  =3    and when x = 1, y = -1....so....the graph must cross the x axis between x = 0 and x = 1.....so...we have another root  on this interval

Finally.....at x = 1, y =-1  and at x = 2, y =4   ...so.....the graph must again cross the  x axis ......so...we have another root

Lastly  at x = x, y= 4  and at x =3, y = -1...so....the graph crosses the x axis one last time...so....we have one more root

So.... since the function crosses the x axis 4 times....we have 4 roots...so....the degree is 4

We have  one negative zero   at x  = -1

And the other three roots are positive