Polynomials question help! (1 Viewer)

[Munkiie]

I already found part a and having trouble with part b of this question, please help.

a) the polynomial equation p(x)=0 has a triple root at x=a. By writing p(x)=(x-a)^3.Q(x), show that p(a)=p'(a)=0.
b) hence find values of u and v if x^5 -9x^4 +ux^3 -18x^2 +v +27 =0 has a triple root at x=3.

Thanks in advance

 

[HeroicPandas]

In part a), should it say "show that: P(a) = P'(a) = P''(a)"?

 

[Munkiie]

No it's just p(a) = p'(a) = 0

 

[asianese]

The condition for a triple root at x=a is that .

 

[TheMathStudent]

Hey,

The trick is to differentiate and sub in 3.

so P'(x)=5x^4-36x^3+3ux^2-36x
sub in 3: P'(3)=0 ... subbing this in you will find that u=25
now do the same for P(3)=0, you will find v=-54.
You didn't need to use p''(3) for this Qn.

hope this helped.