the polynomial p(t)=4t^2-st^2+7 and q(t)=t^2+st+8 leave the same remainder when divided by (t-1) find the value of s
Answers
Answer:
p(t) = 4t^2-st^2+7 , q(t) = t^2+st+8
According to remainder theorem if we divide any polynomial p(x) of degree greater than 1 by any linear polynomial (x+1) then the remainder obtained is p(x)
Here t-1=0
So t = 0-1 = -1
Substituting value of t in p(t)
p(x) = 4t^2-st^2+7
p(-1) = 4(-1)^2 - s(-1)^2 + 7
= 4-s+7
= 11+s
Substituting value of t in q(t)
q(t) = t^2+st+8
q(-1) = (-1)^2 + s(-1) + 8
= 1-s +8
= 9-s
When these polynomials p(t) and q(t) are divided by (t-1) they leave the same remainder
So 11+s = 9-s
( Transposing similar terms on same side )
11-9 = -s-s
2 = -2s
s = -2/2 = -1
Verification
Putting value of s in polynomials and checking if the remainders are same
p(t) = 4t^2-st^2+7
p(-1) = 4(-1)^2-(-1)(-1) + 7
= 4-1+7
= 10
q(t) = t^2+st+8
q(-1) = (-1)^2 + (-1)(-1) + 8
= 1+1+8
= 10
( Hence verified )
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