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Solving fourth degree polynomials as quadratic expressions

Not all fourth degree polynomials can be solved as quadratic expressions, but some fourth degree polynomials can be rewritten as quadratic polynomials and then solved accordingly. For example
`x^4 + 4x^2 - 5`
can be solved as a quadratic expression:

In the above polynomial, let `y = x^2`, then the polynomial becomes,
`y^2 + 4y - 5`
Now we have a quadratic expression that can be solved by factoring or by the quadratic formula. On factoring the above quadratic expression,
`y^2 + 5y - y - 5`
`y(y +5) -1(y + 5)`
`(y + 5)(y - 1)`
Since we are solving for the zeros of the polynomial, applying the zero product rule,
`(y + 5)(y - 1) = 0`
`y + 5 = 0, y - 1 = 0`
`y = -5, y = 1`
Now put back `y = x^2`, so
`x^2 = -5, x^2 =1`
Taking square roots on both sides to get the value of x,
`x = +/- sqrt{-5}, x = +/- sqrt{1}`
Square roots of negative numbers are non real complex numbers written in 'i',  hence the zeros to the above fourth degree polynomial are
`x = {sqrt{5}i, -sqrt{5}i, 1, -1}`

1 comment:

  1. The fourth degree polynomial are solved by converting them in second degree polynomials or we can say in quadratic polynomials.

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