Another Polynomial Question (1 Viewer)

[j1mmy_]

Need help to Find the value of a and b if x = 1 is a double root of x⁴+ax³+bx²-5x+1=0

 

[Carrotsticks]

If x=1 is a double root, then P(1) = 0 and P'(1) = 0 as well, where P'(x) is the first derivative of the polynomial P(x).

Once you sub both of those conditions in, you get a set of simultaneous equations to solve.

 

[superSAIyan2]

You need to use the multiple root theorem
So x =1 is double root of P(x) = x⁴+ax³+bx²-5x+1 and single root of its derivative P'(x) = 4x^3 + 3ax^2 + 2xb - 5
Hence 1 + a + b - 5 +1 = 0 -----> a+b = 3 (1)
Hence 4 + 3a + 2b - 5 = 0 -----> 3a+2b = 1 (2)
Solve simultaneously (1)x3 -(2) : b = 8 and a = -5

 

[j1mmy_]

If x=1 is a double root, then P(1) = 0 and P'(1) = 0 as well, where P'(x) is the first derivative of the polynomial P(x).

Once you sub both of those conditions in, you get a set of simultaneous equations to solve.

You need to use the multiple root theorem
So x =1 is double root of P(x) = x⁴+ax³+bx²-5x+1 and single root of its derivative P'(x) = 4x^3 + 3ax^2 + 2xb - 5
Hence 1 + a + b - 5 +1 = 0 -----> a+b = 3 (1)
Hence 4 + 3a + 2b - 5 = 0 -----> 3a+2b = 1 (2)
Solve simultaneously (1)x3 -(2) : b = 8 and a = -5

Thanks guys However is the multiple root theorem part of the 4 Unit course, since I've never heard of it before? If so, is there a different method to solve this?

 

[superSAIyan2]

yh it is but my class learnt it in three unit for fun lol

 

[Carrotsticks]

You can do the problem (a lot more messy however) without Multiple Root Theorem. Since x=1 is a double root, we know that (x-1)^2 is a factor ie: x^2 - 2x + 1 is a factor. You can use poynomial long division as usual and acquire some remainder in terms of A, B and x. You can then substitute x=1 into the remainder and let that expression be equal to 0, since x=1 is a root. This will be your second equation.

The first equation is found by having P(1) = 0 as usual.