Question

How do you factor completely `P(k)=k^5+3k^4-5k^3-15k^2+4k+12`?

Answer 1

`color(green)((x-2)(x-1)(x+1)(x+2)(x+3))`

Explanation:

Using the Rational Factor Theorem we know that any solutions to
`color(white)("XXX")color(red)1k^5+3k^4-5k^3-15k^2+4k+color(blue)(12)`
will be factors of `color(blue)(12)/(color(red)1)`

That is rational zeros of the given function will be
`color(white)("XXX")k in {+-12,+-6,+-4,+-3,+-2,+-1}`

Evaluating `P(k)` for these values (I used a spreadsheet):

gives us 5 distinct zeros: `k in {2,1,-1,-2,-3}`

Since `P(k)` is a polynomial of degree `5`, it can have at most `5` distinct zeros; so this method has given us all solutions.

The factors are therefore:
`color(white)("XXX")(x-2)(x-1)(x-(-1))(x-(-2))(x-(-3))`

`color(white)("XXX")=(x-2)(x-1)(x+1)(x+2)(x+3)`

Answer 2

`P(k)=(k+3)(k+1)(k-1)(k+2)(k-2)`.

Explanation:

We find that, the sum of the co-effs. is `0," so "(k-1)` is a factor.

Again, the sum of the co-effs. of the even-powered terms

(i.e., `3-15+12=0`) & that of odd-powered (i.e., `1-5+4=0`)

are equal.

`:. (k+1)" is a factor"`.

Thus, `(k^2-1) | P(k).`

Now, `P(k)=ul(k^5+3k^4)-ul(5k^3-15k^2)+ul(4k+12)`

`=k^4(k+3)-5k^2(k+3)+4(k+3)`

`=(k+3)(k^4-5k^2+4)`

Now, knowing that `(k^2-1)` is a factor, we have,

`P(k)=(k+3)(k^4-k^2-4k^2+4)`

`=(k+3){k^2(k^2-1)-4(k^2-1)}`

`=(k+3)(k^2-1)(k^2-4)`

`:. P(k)=(k+3)(k+1)(k-1)(k+2)(k-2)`.

Enjoy Maths.!