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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)`
Posted by sciencesolve on May 1, 2012 at 6:04 AM (Answer #1)
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It has no real roots; it cannot be factored. If graphed, the parabola is above the x-axis....so...no real roots
Posted by elekzy on May 1, 2012 at 2:50 AM (Answer #2)
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