1. `3x^2+6x-1`, find Zeros of a polynomial

Solution:
The Given Polynomial `=3x^2+6x-1`

`3x^2+6x-1=0`

`=>3x^2+6x-1 = 0`

factor is not possible for equation `3x^2+6x-1=0`

But we are trying find solution using the method of perfect square.

Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`

we get, `a=3, b=6, c=-1.`

`:. Delta=b^2-4ac`

`=(6)^2-4 (3) (-1)`

`=36+12`

`=48`

`:. sqrt(Delta)=sqrt(48)=4sqrt(3)`



Now, `alpha=(-b+sqrt(Delta))/(2a)`

`=(-(6)+4sqrt(3))/(2*3)`

`=(-6+4sqrt(3))/6`

`=(-3+2sqrt(3))/3`



and, `beta=(-b-sqrt(Delta))/(2a)`

`=(-(6)-4sqrt(3))/(2*3)`

`=(-6-4sqrt(3))/6`

`=(-3-2sqrt(3))/3`


`=>x = (-3+2sqrt(3))/3" or "x = (-3-2sqrt(3))/3`

---

2. `x^2+3x-4`, find Zeros of a polynomial

Solution:
The Given Polynomial `=x^2+3x-4`

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

`=>x^2+3x-4 = 0`

`=>x^2-x+4x-4 = 0`

`=>x(x-1)+4(x-1) = 0`

`=>(x-1)(x+4) = 0`

`=>(x-1) = 0" or "(x+4) = 0`

`=>x = 1" or "x = -4`

---

3. `2x^2-3x+1`, find Zeros of a polynomial

Solution:
The Given Polynomial `=2x^2-3x+1`

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

`=>2x^2-3x+1 = 0`

`=>2x^2-x-2x+1 = 0`

`=>x(2x-1)-1(2x-1) = 0`

`=>(2x-1)(x-1) = 0`

`=>(2x-1) = 0" or "(x-1) = 0`

`=>2x = 1" or "x = 1`

`=>x = 1/2" or "x = 1`

---

4. `x^3-2x^2-x+2`, find Zeros of a polynomial

Solution:
The Given Polynomial `=x^3-2x^2-x+2`

`x^3-2x^2-x+2=0`

`=>x^3-2x^2-x+2 = 0`

`=>x^3-x^2-x^2+x-2x+2 = 0`

`=>x^2(x-1)-x(x-1)-2(x-1) = 0`

`=>(x-1)(x^2-x-2) = 0`

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

`=>(x-1)(x(x+1)-2(x+1)) = 0`

`=>(x-1)(x+1)(x-2) = 0`

`=>(x-1) = 0" or "(x+1) = 0" or "(x-2) = 0`

`=>x = 1" or "x = -1" or "x = 2`

---