(x + 4/x)² + 36/x + 9x + 20 = 0I'm helping my son with his maths, and it's been a long time since I was able to answer this. Can someone help me, and explain it? I've read some of the previous...

(x + 4/x)² + 36/x + 9x + 20 = 0

I'm helping my son with his maths, and it's been a long time since I was able to answer this. Can someone help me, and explain it?

I've read some of the previous questions and answers, but do not understand the term 'AP' used.

Thanks

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Solve `(x+4/x)^2+36/x+9x+20=0` :

(1) Expand the binomial square:

`(x+4/x)^2+36/x+9x+20=0`

`x^2+8+16/x^2+36/x+9x+20=0`

(2) Multiply through by `x^2` to get rid of fractions:

`x^4+8x^2+16+36x+9x^3+20x^2=0` Collecting terms:

`x^4+9x^3+28x^2+36x+16=0`

(3) If there are any rational zeros, they are of the form `p/q` where `p` is a factor of 16 and `q` is a factor of 1. Thus the possible rational zeros are `(+-1,+-2,+-4,+-8,+-16)`

(4) You can use polynomial long division or synthetic division to find linear factors with remainder zero, or direct substitution to find values of x for which `f(x)=0` .

(By Descarte's rule of signs there can be no positive zeros)((See reference))

Using synthetic division we get:

-1 | 1  9  28  36  16
------------------
1  8  20  16  0

Thus we have `(x+1)(x^3+8x^2+20x+16)=0`

Using synthetic division on the right hand polynomial with divisor -2 yields:

-2 | 1  8  20  16
--------------
1  6  8   0

So we now have `(x+1)(x+2)(x^2+6x+8)=0` . The trinomial factors:

`(x+1)(x+2)(x+2)(x+4)=0`

Thus the real roots are -1,-2,-4 with -2 a repeated root.