# If y = x^2, then find the zeros of the polynomial P(x) = x^2+y^2+2xy+1

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### 2 Answers

Find the zeros of `P(x)=x^2+y^2+2xy+1` if `y=x^2` :

Substitute for `y` to get:

`P(x)=x^2+(x^2)^2+2x(x^2)+1`

`P(x)=x^4+2x^3+x^2+1`

`P(x)=x^2(x^2+2x+1)+1`

`P(x)=x^2(x+1)^2+1`

Setting `P(x)=0` we solve for `x` :

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

`x^2(x+1)^2=-1`

`x(x+1)=+-i` (Take square root of both sides; `i=sqrt(-1)` )

`x^2+x+-i=0` Using the quadratic formula we get:

`x=(-1+-sqrt(1+-4i))/2=1/2(-1+-sqrt(1+-4i))`

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**P(x) has no real zeros. The complex zeros are:**

`1/2(-1+sqrt(1+4i))`

`1/2(-1+sqrt(1-4i))`

`1/2(-1-sqrt(1+4i))`

`1/2(-1-sqrt(1-4i))`

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The problem provides the information that `y = x^2` , hence, you need to substitute `x^2` for y in polynomial P(x) such that:

`P(x) = x^2 + (x^2)^2 + 2x*x^2 + 1`

`P(x) = x^4 + 2x^3 + x^2 + 1`

You need to find the zeroes of polynomial, hence, you should solve the equation `P(x) = 0` such that:

`x^4 + 2x^3 + x^2 + 1 = 0`

Notice that the expression `x^4 + 2x^3 + x^2 + 1` will be larger than zero for any value of x.

**Hence, evaluating the zeroes of the given polynomial yields that P(x) has no real zeroes.**

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