X2 – 1 is a factor of px4 + qx3 + rx2 + sx + t, how that p + r + t = q + s = 0.
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Answered by
67
Let's take
px^4 + qx^3 + rx^2 + sx + t as p (x)
x^2 -1
x^2 = 1
x= +1 or -1
By taking the value as -1
And by substituting the value of x in polynomial p (x)
p (-1) = p -q +r -s +t =0
= p+r+t = q+s
px^4 + qx^3 + rx^2 + sx + t as p (x)
x^2 -1
x^2 = 1
x= +1 or -1
By taking the value as -1
And by substituting the value of x in polynomial p (x)
p (-1) = p -q +r -s +t =0
= p+r+t = q+s
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Answered by
35
x[tex] x^{2} - 1 = 0
=\ \textgreater \ x = +1 or -1[/tex]
Putting x = 1, p(1) = p + q+ r + s + t =0 -----------> i
That being said, put p(-1) = p - q + r -s + t = 0 -------------> ii
Now from ii,
p + r + t = s + q
Also from i,
p + r + t = -s - q
This gives s + q = p + r + t = 0
_____________________________________________________________
Hope it satisfies........xD
Putting x = 1, p(1) = p + q+ r + s + t =0 -----------> i
That being said, put p(-1) = p - q + r -s + t = 0 -------------> ii
Now from ii,
p + r + t = s + q
Also from i,
p + r + t = -s - q
This gives s + q = p + r + t = 0
_____________________________________________________________
Hope it satisfies........xD
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