Year 11 Ext Math 1 Question (1 Viewer)

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

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

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

Gtsh said:

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

Click to expand...

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

Octogonagal said:

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

Click to expand...

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)

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

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]