# (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

### 2 Answers | Add Yours

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.**

what year is this pls?