How do you factor completely P(k)=k^5+3k^4-5k^3-15k^2+4k+12P(k)=k5+3k45k315k2+4k+12?

2 Answers
Jan 9, 2017

color(green)((x-2)(x-1)(x+1)(x+2)(x+3))(x2)(x1)(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)XXX1k5+3k45k315k2+4k+12
will be factors of color(blue)(12)/(color(red)1)121

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

Evaluating P(k)P(k) for these values (I used a spreadsheet):
enter image source here
gives us 5 distinct zeros: k in {2,1,-1,-2,-3}k{2,1,1,2,3}

Since P(k)P(k) is a polynomial of degree 55, it can have at most 55 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))XXX(x2)(x1)(x(1))(x(2))(x(3))

color(white)("XXX")=(x-2)(x-1)(x+1)(x+2)(x+3)XXX=(x2)(x1)(x+1)(x+2)(x+3)

Jan 9, 2017

P(k)=(k+3)(k+1)(k-1)(k+2)(k-2)P(k)=(k+3)(k+1)(k1)(k+2)(k2).

Explanation:

We find that, the sum of the co-effs. is 0," so "(k-1)0, so (k1) is a factor.

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

(i.e., 3-15+12=0315+12=0) & that of odd-powered (i.e., 1-5+4=015+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.!