Question

How do you factor `7x ^ { 9} - 9x + 6x + 4x ^ { 8}`?

Answer 1

Rational factorisation:

`7x^9-9x+6x+4x^8`

`= x(x+1)(7x^7-3x^6+3x^5-3x^4+3x^3-3x^2+3x-3)`

There is one more irrational real zero and three complex conjugate pairs of zeros.

Explanation:

Given:

`7x^9-9x+6x+4x^8`

This example looks a little random, as if containing typos or just made up, but it is interesting to look at for some general principles.

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Standard form

Let us combine like terms and rearrange in standard, descending order of degrees:

`7x^9-9x+6x+4x^8 = 7x^9-3x+4x^8`

`color(white)(7x^9-9x+6x+4x^8) = 7x^9+4x^8-3x`

Note that all of the terms are divisible by `x`, so we can separate that out as a factor:

`color(white)(7x^9-9x+6x+4x^8) = x(7x^8+4x^7-3)`

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Descartes' Rule of Signs

Note at this stage that the pattern of signs of the coefficients of the remaining octic (i.e. degree 8) polynomial is `+ + -`. With one change of sign, this means that `7x^8+4x^7-3` has exactly one positive real zero. Reversing the signs on the term of odd degree, we get the pattern `+ - -`. With one change of sign, we can tell that this octic has exactly one negative real zero. The remainder of the zeros are three complex conjugate pairs of non-real zeros.

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Sum of coefficients shortcut

Next note that `7-4-3 = 0`. Hence `x=-1` is a zero and `(x+1)` a factor:

`7x^9-9x+6x+4x^8`

`= x(x+1)(7x^7-3x^6+3x^5-3x^4+3x^3-3x^2+3x-3)`

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Rational roots theorem

Any rational zeros (and thus factors with rational coefficients) of the remaining septic are expressible in the form `p/q` where `p` is a divisor of the constant term `-3` and `q` a divisor of the coefficient `7` of the leading term.

In addition we know that any real zero is positive, so the only possible rational zeros are:

`1/7, 3/7, 1, 3`

None of these work, so the septic has no rational zeros.

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Where do we go from here?

The remaining septic (i.e. degree 7) polynomial has exactly one real, irrational zero at about `x~~0.843`, but (typically for a polynomial of degree `>= 5`) it is not expressible in terms of radicals. If you like, you can call this zero `alpha` and long divide the septic factor by `(x-alpha)` to get a factorisation in terms of coefficients expressed in terms of `alpha`, but it gets rather messy.