Ext. Maths - Polynomial Question. Please help! :D (1 Viewer)

[mawissah]

The polynomial f(x) is given by f(x) = px³ + 16x² + qx - 120, where p and q are constants. The three zeroes of f(X) are (-2), 3 and a. The value of a is?

I'm not sure how to start. and what does 'zeroes' mean? How do work this out?

It was a multiple choice so the answer is either, 5/9, 4, (-5) or 5

 

[Kurosaki]

Zeroes are the term when u sub in a value of x, and u get P(x)=0. therefore u sub in -2 and -3 to solve for p and q- you know that when u do this, f(x)=0. after finding p and q, sub in a, and solve for a .
Answer is -5

 

[mawissah]

do i find out p and q by simultaneious equations?
thanks ^^

 

[Kurosaki]

f(-2)=0, and f(3)=0. After this, u get

-8p +64-2q-120=0

and

27p+144+3q-120=0.

Rearrange one of these two equations to get one variable in terms of the other, and sub it into the other equation.

e.g
-8p +64-2q-120=0
turns to
-8p-56=2q
-4p-28=q.
sub q=-4p-28 into 27p+144+3q-120=0. from there, you solve for p and q, then sub x=a into f(x) to get f(a), which must equal zero (it is given that a is a zero of f(x))

 

[mawissah]

so i got
p=4
q=-44

and then i substituted a which got
f(a)= 4a^3+16a^2-44a-120

how do i get to find out what a equals, since there are no like terms?

 

[Kurosaki]

well, you have those multiple choice answers in front of you, sub them in until u get a value equal to zero .
This is because it is given that a is a zero of f(x). Hence f(a)=0.

 

[mawissah]

you're the best. i actually love you ^^

 

[Kurosaki]

Nawgh fanks bbby