Hi, I have this problem solved to this:

(x^2-5x+2)^2-(x^2-5x)=8

(x^2-5x+2)(x^2-5x+2)-x^2+5x-8=0

x^4-5x^3+2x^2-5x^3+25x^2-10x+2x^2-10x+4-x^2+5x-8=0

x^4-10x^3+28x^2-15x-4

I am not sure if this problem should (Can) go further?...Please help me?

Guest Nov 4, 2015

#3**+30 **

**Hi, I have this problem solved to this:**

**(x^2-5x+2)^2-(x^2-5x)=8**

**\(\small{ \begin{array}{rcll} (x^2-5x+2)^2-(x^2-5x)&=& 8 \qquad & |\qquad \text{we substitute } z = x^2-5x \\ \\ \hline \\ (z+2)^2 - z &=& 8 \\ z^2+4z+4-z &=& 8 \\ z^2 +3z - 4 &=& 0 \qquad & |\qquad \text{factorisation }\\ (z-1)(z+4) &=& 0 \\ z = 1 &\text{or}& z = -4\\ \\ \hline \\ x^2-5x &=& z \qquad & |\qquad z = -4 \\ x^2-5x &=& -4 \\ x^2-5x+4 &=& 0 \qquad & |\qquad \text{factorisation }\\ (x-1)(x-4) &=& 0 \\ x = 1 &\text{or}& x = 4\\ \\ \hline \\ x^2-5x &=& z \qquad & |\qquad z = 1 \\ x^2-5x &=& 1 \\ x^2-5x-1 &=& 0\\ \boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ x^2-5x-1 &=& 0 \qquad & |\qquad a=1 \quad b=-5 \quad c=-1\\ x &=& {5 \pm \sqrt{25-4 \cdot (-1)} \over 2}\\ x &=& {5 \pm \sqrt{29} \over 2}\\ x = {5 + \sqrt{29} \over 2} &\text{or}& x = {5 - \sqrt{29} \over 2} \\ x = 5.1925824036 &\text{or}& x = -0.1925824036 \\ \end{array} }\)**

heureka Nov 4, 2015

#1**+10 **

\((x^2-5x+2)^2-(x^2-5x)=8\\ (x^2-5x+2)(x^2-5x+2)-x^2+5x-8=0\\ x^4-5x^3+2x^2-5x^3+25x^2-10x+2x^2-10x+4-x^2+5x-8=0\\ ...\\ x^4-10x^3+28x^2-15x-4=0\\ \)

Now I am going to look for factors. I am using the remainder theorem.

If I sub in x=1, which was just the first number that I tried. I get

1 - 10 +28 - 15 - 4 and that DOES **=0 ** so that means that one of the factors is (x-1)

so now I am going to do an algebraic division to see what (x-1) would be multiplied by

x^3 - 9x^2 + 19x + 4

---------------------------------------------

x-1 | x^4 - 10x^3 + 28x^2 - 15x - 4

x^4 - x^3

--------------------------

- 9x^3+28x^2

- 9x^3 +9x^2

------------------------

19x^2 - 15x

19x^2 - 19x

---------------------

4x - 4

4x - 4

--------

0

-------

So we have

\((x-1)(x^3-9x^2+19x+4)=0\)

I tested +1,-1,+2,-2,+4,-4 I chose these because they are all factors of 4

x=4 made that second bracket = 0 therefore another factor is (x-4)

I did another algebraic division the same as before and found that

\((x^3-9x^2+19x+4)\div (x-4)=x^2-5x-1\\ so\\ (x-1)(x-4)(x^2-5x-1)=0\\ so\\ x-1=0\qquad or \qquad x-4=0 \qquad or\qquad x^2-5x-1=0\\ x=1\qquad or \qquad x=4 \qquad or\qquad x=\frac{5\pm\sqrt{25-\;-4}}{2*1}\\ x=1\qquad or \qquad x=4 \qquad or\qquad x=\frac{5\pm\sqrt{29}}{2}\\ x=1\quad or \quad x=4 \quad or\quad x=\frac{5+\sqrt{29}}{2}\quad or \quad x=\frac{5-\sqrt{29}}{2} \)

\(x=1 \quad or \quad x=4 \quad or \quad x\approx5.193 \quad or \quad x\approx -0.193\)

NOW I have not checked thes BUT if you substitute them back into the initial equation they should all make the equation true.

Let's have a look at the graph y=(x^2-5x+2)^2-(x^2-5x)-8 and see where y = 0

https://www.desmos.com/calculator/lhcjfgppxk

There you go, that is how it is done. If you would like to ask questions go ahead and ask. :))

Melody Nov 4, 2015

#2**+5 **

Hi Melody,

Thank you sooo much for this, however I do seem to have a teeny problem, which is:

1-10+28-15-4 I get to be 0?

The above is:

=(1-10)+28-15-4

=(-9+28)-15-4

=(19-15)-4

=0

This would change the rest of the sum considerably, do you agree?.. :-)

Guest Nov 4, 2015

#3**+30 **

Best Answer

**Hi, I have this problem solved to this:**

**(x^2-5x+2)^2-(x^2-5x)=8**

**\(\small{ \begin{array}{rcll} (x^2-5x+2)^2-(x^2-5x)&=& 8 \qquad & |\qquad \text{we substitute } z = x^2-5x \\ \\ \hline \\ (z+2)^2 - z &=& 8 \\ z^2+4z+4-z &=& 8 \\ z^2 +3z - 4 &=& 0 \qquad & |\qquad \text{factorisation }\\ (z-1)(z+4) &=& 0 \\ z = 1 &\text{or}& z = -4\\ \\ \hline \\ x^2-5x &=& z \qquad & |\qquad z = -4 \\ x^2-5x &=& -4 \\ x^2-5x+4 &=& 0 \qquad & |\qquad \text{factorisation }\\ (x-1)(x-4) &=& 0 \\ x = 1 &\text{or}& x = 4\\ \\ \hline \\ x^2-5x &=& z \qquad & |\qquad z = 1 \\ x^2-5x &=& 1 \\ x^2-5x-1 &=& 0\\ \boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ x^2-5x-1 &=& 0 \qquad & |\qquad a=1 \quad b=-5 \quad c=-1\\ x &=& {5 \pm \sqrt{25-4 \cdot (-1)} \over 2}\\ x &=& {5 \pm \sqrt{29} \over 2}\\ x = {5 + \sqrt{29} \over 2} &\text{or}& x = {5 - \sqrt{29} \over 2} \\ x = 5.1925824036 &\text{or}& x = -0.1925824036 \\ \end{array} }\)**

heureka Nov 4, 2015

#4**+5 **

Sorry for confusing you guest. That was a small mistype. It was =0 not =1

I have fixed it in my answer. My answer is correct. :))

I like Heureka's answer better though.

Thanks Heureka :))

Melody Nov 4, 2015

#5**+5 **

No problem Melody. Thank you very much indeed for this solution. Also Heureka, Thank you, thank you and again..a big thank you. wow!. Just something, how do you get the squares to show "normal", as opposed to the "^" character?..

Guest Nov 4, 2015

#6**+5 **

You can get superscript and subscript formats using the ribbon.

But Heurka and I both use LaTex to write our math speak in.

The Latex gateway is in the ribbon too.

If you decide you want to have a look at LaTex there is a rather messy thread on it in the sticky notes. Some of our youngest members have learned it in the past but it does take a little effort. :)

Melody Nov 4, 2015

#7**+5 **

Thanx Melody...uhm, just one thing more please, I am batteling a bit to see (understand) the reasoning behind getting the factors (x-1) and (x-4)..I will still try to figure it out, but was wondering if you would be so kind to just take me through that?..If your'e busy or there is a reason why you cannot, (don't want to I mean), I will understand, no hard feelings?

Guest Nov 4, 2015

#8**+5 **

Hi Guest,

I do not know if you will even see this since you aske the question 24 hous ago, if you did not save the thread somewhere you have probably already lost track of it.

SO pleas join up and then you questions will be automatically put into you watchlist and you can aske for email notification if anyone answers you.

Anyway, to learn about polynomial long division it would be good for you to watch this video :))

If you have more questions you better join up or I am not likely to see it.

You can private message me with the address of you post included then I could just jump over to it :))

SO JOIN UP PLEASE!!

Melody Nov 5, 2015