# help me please factor 4x^2-3x+5 PLEASE HELP!!

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You need to remember that you may write the factored form of a polynomial of n-th order such that:

`f(x) = a(x-x_1)(x-x_2)(x-x_3)....(x-x_n)`

`x_1,....x_n` are the roots of polynomial

Hence, you should find the roots of quadratic `4x^2-3x+5 = 0` such that:

`x_(1,2) = (3+-sqrt(9-80))/8`

`x_(1,2) = (3+-sqrt(-71))/8`

You need to remember that the complex number theory tell that `sqrt(-1) = i` .

`x_(1,2) = (3+-i*sqrt(71))/8`

**Hence, you may write the factored form of given quadratic using the conjugate complex roots such that:**

**`4x^2-3x+5 = 4(x - (3+isqrt71)/8)(x -(3-isqrt71)/8)` **

It has no real roots; it cannot be factored. If graphed, the parabola is above the x-axis....so...no real roots