Question

How do you factor the polynomial `x^4-x^3-13x^2-7x-140`?

Answer 1

`x^4-x^3-13x^2-7x-140 = (x-5)(x+4)(x^2+7)`

`color(white)(x^4-x^3-13x^2-7x-140) = (x-5)(x+4)(x-sqrt(7)i)(x+sqrt(7)i)`

Explanation:

Given:

`f(x) = x^4-x^3-13x^2-7x-140`

By the rational root theorem, any rational zeros of this quartic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of `140` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-5, +-7, +-10, +-14, +-20, +-28, +-35, +-70, +-140`

In addition, note that the pattern of signs of the coefficients is `+ - - - -`. With one change of signs, Descartes' Rule of Signs tells us that this quartic has exactly one positive real zero.

So, trying each of the positive rational possibilities in turn, we get to:

`f(5) = (color(blue)(5))^4-(color(blue)(5))^3-13(color(blue)(5))^2-7(color(blue)(5))-140`

`color(white)(f(5)) = 625-125-325-35-140 = 0`

So `x=5` is a zero and `(x-5)` a factor:

`x^4-x^3-13x^2-7x-140 = (x-5)(x^3+4x^2+7x+28)`

Note that in the remaining cubic the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

`x^3+4x^2+7x+28 = (x^3+4x^2)+(7x+28)`

`color(white)(x^3+4x^2+7x+28) = x^2(x+4)+7(x+4)`

`color(white)(x^3+4x^2+7x+28) = (x^2+7)(x+4)`

The remaining quadratic has no real zeros and no factors with real coefficients, but we can factor it with complex coefficients...

`x^2+7 = x^2+(sqrt(7))^2 = x^2-(sqrt(7)i)^2 = (x-sqrt(7)i)(x+sqrt(7i))`

Putting it all together, we have:

`x^4-x^3-13x^2-7x-140 = (x-5)(x+4)(x^2+7)`

`color(white)(x^4-x^3-13x^2-7x-140) = (x-5)(x+4)(x-sqrt(7)i)(x+sqrt(7)i)`