Year 11 Ext Math 1 Question (1 Viewer)

[Octogonagal]

Question is; A polynomial P(x) of degree 4 is known to have zeroes 2 and -2

a) Write an expression for P(x)
b) Given that P(0)=4 and P(1)=-3, find a further expression for P(x)
c) Solve P(x) = 0

No one has any idea how to do it

 

[d1zzyohs]

maybe i'm dumb; but i think there's not enough information.

 

[Gtsh]

Could u do P(X)=(x-2)(x+2)(x+a)(x+b)
Sub is P(0) and P(1) to get two simultaneous equations to find the values for a and b
Not sure if it will work though

 

[d1zzyohs]

Could u do P(X)=(x-2)(x+2)(x+a)(x+b)
Sub is P(0) and P(1) to get two simultaneous equations to find the values for a and b
Not sure if it will work though

You would need the extra info that the polynomial is monic - which is why I said we need more info.
If this was given; your method would work

 

[Deem_Skills]

Question is; A polynomial P(x) of degree 4 is known to have zeroes 2 and -2

a) Write an expression for P(x)
b) Given that P(0)=4 and P(1)=-3, find a further expression for P(x)
c) Solve P(x) = 0

No one has any idea how to do it

a) P(x) = a(x+2)(x-2)(x+b)(x+c)
not sure for b and c cause like d1zzy mentioned there isnt enough info to get a b and c (i think)

 

[fx82au]

yeah defo missing some information. does it mention if there are any double roots?

 

[Unknown_Xp998]

Given that the polynomial is degree 4, without loss of generality we can assume that it is monic, as the leading coefficient has no bearing on the roots of P(x):

a) P(x) = (x-2)(x+2)(x^2 +bx + c) [Since we know that the polynomial is degree 4]
b) Now P(0) = (0-2)(0+2)(0^2 + b(0) + c) = 4
(-4)(c) = 4
c = -1

And P(1) = (1-2)(1+2)(1^2+b(1) + c) = -3
(-3)(1 + b -1) = -3
b = 1
P(x) = (x-2)(x+2)(x^2 + x -1)

c) P(x) = (x-2)(x+2)(x^2 + x -1) = 0
x = ±2, (-1 ± sqrt5)/2 [from solving x^2 + x -1 = 0 via the quadratic formula]