Question

`2x^4-9x^3-7x^2+54x-40=0`?

Answer 1

The roots are `1, 2, 4, -5/2`

Explanation:

Given:

`2x^4-9x^3-7x^2+54x-40=0`

By the rational roots theorem, any rational roots of this polynomial equation must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-40` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational roots are:

`+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-8, +-10, +-20, +-40`

There is a shortcut to trying these in that the sum of the coefficients of the given quartic is zero. That is:

`2-9-7+54-40 = 0`

We can deduce that `x=1` is a root and `(x-1)` a factor:

`2x^4-9x^3-7x^2+54x-40 = (x-1)(2x^3-7x^2-14x+40)`

Notice that the ratio of the first and second terms of the remaining cubic is different from that of the third and fourth terms. So this cubic will not factor by grouping.

Let's try one of the possible rational roots, `2`:

`2(color(blue)(2))^3-7(color(blue)(2))^2-14(color(blue)(2))+40 = 16-28-28+40 = 0`

So `x=2` is a root and `(x-2)` a factor:

`2x^3-7x^2-14x+40 = (x-2)(2x^2-3x-20)`

We can factor the remaining quadratic using an AC method:

Find a pair of factors of `AC = 2*20 = 40` which differ by `B=3`.

The pair `8, 5` works.

Use this pair to split the middle term and factor by grouping:

`2x^2-3x-20 = (2x^2-8x)+(5x-20)`

`color(white)(2x^2-3x-20) = 2x(x-4)+5(x-4)`

`color(white)(2x^2-3x-20) = (2x+5)(x-4)`

Hence the remaining zeros are:

`x = -5/2" "` and `" "x = 4`