h(x)=x^4+4x^3-33x^2+ax+b
where a and b are constants
i)given that (x+1) and (x-1) are factors of h(x), determine the values of a and b.
ii)fully factorise h(x)
iii)hence solve the equation h(x)=0.
We can use synthetic division
If x + 1 is a factor, then -1 is a root
So
-1 [ 1 4 -33 a b ]
-1 -3 36 -a -36
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1 3 -36 a + 36 b - a - 36 ⇒ b - a - 36 = 0 ⇒ -a + b = 36 (1)
And if x -1 is a factor, the x = 1 is a root
So
1 [ 1 4 -33 a b ]
1 5 -28 a - 28
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1 5 -28 a - 28 a + b - 28 = 0 ⇒ a + b = 28 (2)
Using (1) and (2) we have that
-a + b = 36
a + b = 28 add these
2b = 64
b = 32 so a = -4
If ( x +1) and (x -1) are factors then (x + 1) ( x -1) = (x^2 - 1) is also a factor
Using polynomial division
x^2 + 4x - 32
x^2 - 1 [ x^4 + 4x^3 - 33x^2 - 4x + 32 ]
x^4 - 1x^2
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4x^3 - 32x^2 -4x
4x^3 -4x
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-32x^2 + 32
-32x^2 + 32
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The remaining polynomial x^2 + 4x - 32 factors as (x + 8) ( x - 4)
Full factorization is (x -1) ( x + 1) ( x + 8) (x - 4)
h (x) = 0 when x = 1 or x = -1 or x = -8 or x = 4