Math, asked by s2124sambitkumarsahu, 4 hours ago

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

Answered by js6326742
18

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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